Some Math Connotations Demystified 数学内涵解密

This Taiwanese Math Prof is very approachable in clarifying the doubts in an unconventional way different from the arcane textbook definitions. Below are his few key tips to breakthrough the “mystified”concepts :

1. “Dual Space“(对偶空间) : it is the evaluation of a “Vector Space”.

Example: A student studies few subjects {Math, Physics, English, Chemistry…}, these subjects form a “Subject Vector Space” (V), if we associate the subjects with weightages (加权) , say, Math 4, Physics 3, English 2, Chemistry 1, the “Weightage Dual Space” of V will be W= {4, 3, 2, 1}.

2. Vector: beyond the meaning of a physical vector with direction and value, it extends to any “object” which can be manipulated (抵消) by the 4 operations “+, – , x, / ” in a FieldF = {R or Z2 …}.

Eg. $latex alpha_{1}.v_{1} + alpha_{2}.v_{2} + alpha_{3}.v_{3}, forall alpha_{j} in… View original post 111 more words Advertisements Posted in math | Leave a comment Homology: Why Boundary of Boundary = 0 ? This equation puzzles most people. WHY ?$latex boxed {{delta}^2 = 0 { ?}}&fg=aa0000&s=3 $It is analogous to the Vector Algebra: Let the boundary of {A, B} =$latex delta (A,B) = overrightarrow{AB }latex overrightarrow{AB } + overrightarrow {BA} =overrightarrow{AB } – overrightarrow {AB} = vec 0 $Note: Co-homology: (上)同调 Euclid Geometry & Homology: Isabell Darcy Lecture: cohomology View original post Posted in math | Leave a comment Existence and properties of normal closure If $E$ is an algebraic extension field of $K$, then there exists an extension field $F$ of $E$ (called the normal closure of $E$ over $K$) such that (i) $F$ is normal over $K$; (ii) no proper subfield of $F$ containing $E$ is normal over $K$; (iii) if $E$ is separable over $K$, then $F$ is Galois over $K$; (iv) $[F:K]$ is finite if and only if $[E:K]$ is finite. The field $F$ is uniquely determined up to an $E$-isomorphism. Proof: (i) Let $X=\{u_i\mid i\in I\}$ be a basis of $E$ over $K$ and let $f_i\in K[x]$ be the minimal polynomial of $u_i$. If $F$ is a splitting field of $S=\{f_i\mid i\in I\}$ over $E$, then $F=E(Y)$, where $Y\supseteq X$ is the set of roots of the $f_i$. Then $F=K(X)(Y)=K(Y)$ so $F$ is also a splitting field of $S$ over $K$, hence $F$ is normal over $K$ as it is the splitting field of a family of polynomials in $K[x]$. (iii) If $E$ is separable over $K$, then each $f_i$ is separable. Therefore $F$ is Galois over $K$ as it is a splitting field over $K$ of a set of separable polynomials in $K[x]$. (iv) If $[E:K]$ is finite, then so is $X$ and hence $S$. Say $S=\{f_1,\dots,f_n\}$. Then $F=E(Y)$, where $Y$ is the set of roots of the $f_i$. Then $F$ is finitely generated and algebraic, thus a finite extension. So $[F:K]$ is finite. (ii) A subfield $F_0$ of $F$ that contains $E$ necessarily contains the root $u_i$ of $f_i\in S$ for every $i$. If $F_0$ is normal over $K$ (so that each $f_i$ splits in $F_0$ by definition), then $F\subset F_0$ (since $F$ is the splitting field) and hence $F=F_0$. Finally let $F_1$ be another extension field of $E$ with properties (i) and (ii). Since $F_1$ is normal over $K$ and contains each $u_i$, $F_1$ must contain a splitting field $F_2$ of $S$ over $K$ with $E\subset F_2$. $F_2$ is normal over $K$ (splitting field over $K$ of family of polynomials in $K[x]$), hence $F_2=F_1$ by (ii). Therefore both $F$ and $F_1$ are splitting fields of $S$ over $K$ and hence of $S$ over $E$: If $F=K(Y)$ (where $Y$ is set of roots of $f_i$) then $F\subseteq E(Y)$ since $E(Y)$ contains $K$ and $Y$. Since $Y\supseteq X$, so $K(Y)$ contains $E=K(X)$ and $Y$, hence $F=E(Y)$. Hence the identity map on $E$ extends to an $E$-isomorphism $F\cong F_1$. Posted in math | Tagged | Leave a comment Happy New Year to Readers of Mathtuition88.com Wishing all readers of Mathtuition88.com a happy new year, and may 2017 bring you peace and joy in your life. No matter which stage of life you are in (student/career/parent/retiree), here is my sincere wishes that you will achieve your goals in 2017, and more importantly be happy in the process. Best wishes, Mathtuition88.com Posted in math | Tagged | Leave a comment Cours Raisonnements (Logics) , Ensembles ( Sets), Applications (Mappings) This is an excellent quick revision of the French Baccalaureat Math during the first month of French university. (Unfortunately common A-level Math syllabus lacks such rigourous Math foundation.) Most non-rigourous high-school students / teachers abuse the use of : “=> ” , “<=>” . Prove by “Reductio par Absudum” 反证法 (by Contradiction) is a clever mathematical logic :$latex boxed {(A => B) <=> (non B => non A)} &s=3$Famous Examples: 1) Prove$latex sqrt 2 is irrational ; 2) There are infinite prime numbers (both by Greek mathematician Euclid 3,000 years ago) The young teacher showed the techniques of proving Mapping: Surjective (On-to) – best understood in Chinese 满射 (Full Mapping) Injective (1-to-1) 单射 Bijective (On-to & 1-to-1) 双射 He used an analogy of (the Set of) red Indians shooting (the Set of bisons 野牛): All bisons are shot by arrows from1 or more Indians. (Surjective… View original post 81 more words Posted in math | Leave a comment Printable Calendar 2017 2017 Calendar Printable Calendar 2017, with (Singapore) holidays. Generated by http://www.calendarlabs.com/customize/pdf-calendar/monthly-calendar-01. Posted in math | Tagged | Leave a comment A little more perseverance, maybe success is near So close yet so far, to the heap of diamonds… Wishing all readers a happy new year ahead. May your dreams and wishes come true! 再努力一下，或许就是成功! Posted in math | Tagged | Leave a comment A finitely generated torsion-free module A over a PID R is free A finitely generated torsion-free module $A$ over a PID $R$ is free. Proof (Hungerford 221) If $A=0$, then $A$ is free of rank 0. Now assume $A\neq 0$. Let $X$ be a finite set of nonzero generators of $A$. If $x\in X$, then $rx=0$ ($r\in R$) if and only if $r=0$ since $A$ is torsion-free. Consequently, there is a nonempty subset $S=\{x_1,\dots,x_k\}$ of $X$ that is maximal with respect to the property: $\displaystyle r_1x_1+\dots+r_kx_k=0\ (r_i\in R) \implies r_i=0\ \text{for all}\ i.$ The submodule $F$ generated by $S$ is clearly a free $R$-module with basis $S$. If $y\in X-S$, then by maximality there exist $r_y,r_1,\dots,r_k\in R$, not all zero, such that $r_yy+r_1x_1+\dots+r_kx_k=0$. Then $r_yy=-\sum_{i=1}^kr_ix_i\in F$. Furthermore $r_y\neq 0$ since otherwise $r_i=0$ for every $i$. Since $X$ is finite, there exists a nonzero $r\in R$ (namely $r=\prod_{y\in X-S}r_y$) such that $rX=\{rx\mid x\in X\}$ is contained in $F$: If $y_i\in X-S$, then $ry=r_{y_1}\dots r_{y_n}y_i\in F$ since $r_{y_i}y_i\in F$. If $x\in S$, then clearly $rx\in F$ since $F$ is generated by $S$. Therefore, $rA=\{ra\mid a\in A\}\subset F$. The map $f:A\to A$ given by $a\mapsto ra$ is an $R$-module homomorphism with image $rA$. Since $A$ is torsion-free $\ker f=0$, hence $A\cong rA\subset F$. Since a submodule of a free module over a PID is free, this proves $A$ is free. Posted in math | Tagged | Leave a comment Tensor is a right exact functor Elementary Proof This is a relatively elementary proof (compared to others out there) of the fact that tensor is a right exact functor. Proof is taken from Hungerford, and reworded slightly. The key prerequisites needed are the universal property of quotient and of tensor product. Statement If $A\xrightarrow{f}B\xrightarrow{g}C\to 0$ is an exact sequence of left modules over a ring $R$ and $D$ is a right $R$-module, then $\displaystyle D\otimes_R A\xrightarrow{1_D\otimes f}D\otimes_R B\xrightarrow{1_D\otimes g}D\otimes_R C\to 0$ is an exact sequence of abelian groups. An analogous statement holds for an exact sequence in the first variable. Proof (Hungerford 210) We split our proof into 3 parts: (i) $\text{Im}(1_D\otimes g)=D\otimes_R C$; (ii) $\text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g)$; and (iii) $\text{Ker}(1_D\otimes g)\subseteq\text{Im}(1_D\otimes f)$. (i) Since $g$ is an epimorphism by hypothesis every generator $d\otimes c$ of $D\otimes_R C$ is of the form $d\otimes g(b)=(1_D\otimes g)(d\otimes b)$ for some $b\in B$. Thus $\text{Im}(1_D\otimes g)$ contains all generators of $D\otimes_R C$, hence $\text{Im}(1_D\otimes g)=D\otimes_R C$. (ii) Since $\text{Ker} g=\text{Im} f$ we have $gf=0$ and $\displaystyle (1_D\otimes g)(1_D\otimes f)=1_D\otimes gf=1_D\otimes 0=0,$ hence $\text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g)$. (iii) Let $\pi:D\otimes_R B\to(D\otimes_R B)/\text{Im}(1_D\otimes f)$ be the canonical epimorphism. From (ii), $\text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g)$ so (by universal property of quotient Theorem 1.7) there is a homomorphism $\alpha:(D\otimes_R B)/\text{Im}(1_D\otimes f)\to D\otimes_R C$ such that $\displaystyle \alpha(\pi(d\otimes b))=(1_D\otimes g)(d\otimes b)=d\otimes g(b).$ We shall show that $\alpha$ is an isomorphism. Then $\text{Ker}(1_D\otimes g)=\text{Im}(1_D\otimes f)$. We show first that the map $\beta:D\times C\to(D\otimes_R B)/\text{Im}(1_D\otimes f)$ given by $(d,c)\mapsto\pi(d\otimes b)$, where $g(b)=c$, is independent of the choice of $b$. Note that there is at least one such $b$ since $g$ is an epimorphism. If $g(b')=c$, then $g(b-b')=0$ and $b-b'\in\text{Ker} g=\text{Im} f$, hence $b-b'=f(a)$ for some $a\in A$. Since $d\otimes f(a)\in\text{Im}(1_D\otimes f)$ and $\pi(d\otimes f(a))=0$, we have \begin{aligned} \pi(d\otimes b)&=\pi(d\otimes(b'+f(a))\\ &=\pi(d\otimes b'+d\otimes f(a))\\ &=\pi(d\otimes b')+\pi(d\otimes f(a))\\ &=\pi(d\otimes b'). \end{aligned} Therefore $\beta$ is well-defined. Verify that $\beta$ is middle linear: \begin{aligned} \beta(d_1+d_2,c)&=\pi((d_1+d_2)\otimes b)\qquad\text{where }g(b)=c\\ &=\pi(d_1\otimes b+d_2\otimes b)\\ &=\pi(d_1\otimes b)+\pi(d_2\otimes b)\\ &=\beta(d_1,c)+\beta(d_2,c). \end{aligned} \begin{aligned} \beta(d,c_1+c_2)&=\pi(d\otimes(b_1+b_2))\qquad\text{where }g(b_i)=c_i\\ &=\pi(d\otimes b_1+d\otimes b_2)\\ &=\pi(d\otimes b_1)+\pi(d\otimes b_2)\\ &=\beta(d,c_1)+\beta(d,c_2). \end{aligned} Let $r\in R$. \begin{aligned} \beta(dr,c)&=\pi(dr\otimes b)\qquad\text{where }g(b)=c\\ &=\pi(d\otimes rb)\\ &=\beta(d,rc)\qquad\text{where }g(rb)=rg(b)=rc. \end{aligned} By universal property of tensor product there exists a unique homomorphism $\bar{\beta}:D\otimes_R C\to(D\otimes_R B)/\text{Im}(1_D\otimes f)$ such that $\bar{\beta}(d\otimes c)=\beta(d,c)=\pi(d\otimes b)$, where $g(b)=c$. Therefore, for any generator $d\otimes c$ of $D\otimes_R C$, $\displaystyle \alpha\bar{\beta}(d\otimes c)=\alpha\pi(d\otimes b)=d\otimes g(b)=d\otimes c,$ hence $\alpha\bar{\beta}$ is the identity map. Similarly \begin{aligned} \bar{\beta}\alpha(d\otimes b+\text{Im}(1_D\otimes f))&=\bar{\beta}\alpha\pi(d\otimes b)\\ &=\bar{\beta}(d\otimes g(b))\\ &=\pi(d\otimes b)\\ &=d\otimes b+\text{Im}(1_D\otimes F) \end{aligned} so $\bar{\beta}\alpha$ is the identity so that $\alpha$ is an isomorphism. Posted in math | Tagged | Leave a comment Category Theory in Computing Languages Yes, lots. Just one example: a function with 2 inputs from A and B and results from C would have the type A x B -> C but in functional languages like Haskell we are using A -> (B -> C), i.e. a function that returns a function. This “currying” is exactly a the categorical definition of a cartesian closed category as one where Hom(AxB,C) is isomorphic to Hom(A,B -> C) and in this false you can replace Hom(X,Y) with X -> Y. It is well known that effects in functional programming can be modelled by monads which is a concept from category theory. Nowadays a weaker structure called applicative functors has become very popular – needless to say also a concept from Category Theory. Not all languages are functional (yet) but… View original post 24 more words Posted in math | Leave a comment Note on Finitely Generated Abelian Groups We state and prove a sufficient condition for finitely generated Abelian Groups to be the direct product of its generators, and state a counterexample to the conclusion when the condition is not satisfied. Theorem Let $G$ be an abelian group and $G=\langle g_1,\dots, g_n\rangle$. Suppose the generators $g_1,\dots,g_n$ are linearly independent over $\mathbb{Z}$, that is, whenever $c_1g_1+\dots+c_ng_n=0$ for some integers $c_i\in\mathbb{Z}$, we have $c_1=\dots=c_n=0$. (Here we are using additive notation for $(G,+)$, where the identity of $G$ is written as 0, the inverse of $g$ is written as $-g$). Then $\displaystyle G\cong\langle g_1\rangle\times\dots\times\langle g_n\rangle.$ Proof Define the following map $\psi:\langle g_1\rangle\times\dots\times\langle g_n\rangle\to\langle g_1,\dots,g_n\rangle$ by $\displaystyle \psi((c_1g_1,\dots,c_ng_n))=c_1g_1+\dots+c_ng_n.$ We can check that $\psi$ is a group homomorphism. We have that $\psi$ is surjective since any element $x\in\langle g_1,\dots,g_n\rangle$ is by definition a combination of finitely many elements of the generating set and their inverses. Since $G$ is abelian, $x=c_1g_1+\dots+c_ng_n$ for some $c_i\in\mathbb{Z}$. Also, $\psi$ is injective since if $c_1g_1+\dots+c_ng_n=0$, then all the coefficients $c_i$ are zero (by the linear independence condition). Thus $\ker\psi$ is trivial. Hence $\psi$ is an isomorphism. Remark Note that without the linear independence condition, the conclusion may not be true. Consider $G=\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_5$ which is abelian with order 30. Consider $g_1=(1,1,0)$, $g_2=(0,1,1)$. We can see that $G=\langle g_1,g_2\rangle$, by observing that $3g_1=(1,0,0)$, $4g_1=(0,1,0)$, $2g_1+g_2=(0,0,1)$. However $\langle g_1\rangle\times\langle g_2\rangle=\mathbb{Z}_6\times\mathbb{Z}_{15}$ has order 90. Thus $\langle g_1,g_2\rangle\not\cong\langle g_1\rangle\times\langle g_2\rangle$. Posted in math | Tagged | Leave a comment Locally Lipschitz implies Lipschitz on Compact Set Proof Assume $\phi$ is locally Lipschitz on $\mathbb{R}^n$, that is, for any $x\in \mathbb{R}^n$, there exists $\delta, L>0$ (depending on $x$) such that $|\phi(z)-\phi(y)|\leq L|z-y|$ for all $z,y\in B_\delta(x)=\{t\in\mathbb{R}^n: |x-t|<\delta\}$. Then, for any compact set $K\subset\mathbb{R}^n$, there exists a constant $M>0$ (depending on $K$) such that $|\phi(x)-\phi(y)|\leq M|x-y|$ for all $x,y\in K$. That is, $\phi$ is Lipschitz on $K$. Proof Suppose to the contrary $\phi$ is not Lipschitz on $K$, so that for all $M>0$, there exists $x,y\in K$ such that $\displaystyle \frac{|\phi(x)-\phi(y)|}{|x-y|}>M.$ Then there exists two sequences $x_n, y_n\in K$ such that $\displaystyle \frac{|\phi(x_n)-\phi(y_n)|}{|x_n-y_n|}\to\infty.$ Since $\phi$ is locally Lipschitz implies $\phi$ is continuous, so $\phi$ is bounded on $K$ by Extreme Value Theorem. Hence $|x_n-y_n|\to 0$. By sequential compactness of $K$, there exists a convergent subsequence $x_{n_k}\to x$, and thus $y_{n_k}\to x$. Then for any $L>0$, there exists $k$ such that $x_{n_k},y_{n_k}\in B_\delta(x)$ but $\displaystyle \frac{|\phi(x_{n_k})-\phi(y_{n_k})|}{|x_{n_k}-y_{n_k}|}>L$ which contradicts that $\phi$ is locally Lipschitz. Posted in math | Tagged | Leave a comment Gauss Lemma Proof There are two related results that are commonly called “Gauss Lemma”. The first is that the product of primitive polynomial is still primitive. The second result is that a primitive polynomial is irreducible over a UFD (Unique Factorization Domain) D, if and only if it is irreducible over its quotient field. Gauss Lemma: Product of primitive polynomials is primitive If $D$ is a unique factorization domain and $f,g\in D[x]$, then $C(fg)=C(f)C(g)$. In particular, the product of primitive polynomials is primitive. Proof (Hungerford pg 163) Write $f=C(f)f_1$ and $g=C(g)g_1$ with $f_1$, $g_1$ primitive. Consequently $\displaystyle C(fg)=C(C(f)f_1C(g)g_1)\sim C(f)C(g)C(f_1g_1).$ Hence it suffices to prove that $f_1g_1$ is primitive, that is, $C(f_1g_1)$ is a unit. If $f_1=\sum_{i=0}^n a_ix^i$ and $g_1=\sum_{j=0}^m b_jx^j$, then $f_1g_1=\sum_{k=0}^{m+n}c_kx^k$ with $c_k=\sum_{i+j=k}a_ib_j$. If $f_1g_1$ is not primitive, then there exists an irreducible element $p$ in $D$ such that $p\mid c_k$ for all $k$. Since $C(f_1)$ is a unit $p\nmid C(f_1)$, hence there is a least integer $s$ such that $\displaystyle p\mid a_i\ \text{for}\ i Similarly there is a least integer $t$ such that $\displaystyle p\mid b_j\ \text{for}\ j Since $p$ divides $\displaystyle c_{s+t}=a_0b_{s+t}+\dots+a_{s-1}b_{t+1}+a_sb_t+a_{s+1}b_{t-1}+\dots+a_{s-t}b_0,$ $p$ must divide $a_sb_t$. Since every irreducible element in $D$ (UFD) is prime, $p\mid a_s$ or $p\mid b_t$. This is a contradiction. Therefore $f_1g_1$ is primitive. Primitive polynomials are associates in $D[x]$ iff they are associates in $F[x]$ Let $D$ be a unique factorization domain with quotient field $F$ and let $f$ and $g$ be primitive polynomials in $D[x]$. Then $f$ and $g$ are associates in $D[x]$ if and only if they are associates in $F[x]$. Proof ($\impliedby$) If $f$ and $g$ are associates in the integral domain $F[x]$, then $f=gu$ for some unit $u\in F[x]$. Since the units in $F[x]$ are nonzero constants, so $u\in F$, hence $u=b/c$ with $b,c\in D$ and $c\neq 0$. Thus $cf=bg$. Since $C(f)$ and $C(g)$ are units in $D$, $\displaystyle c\sim cC(f)\sim C(cf)=C(bg)\sim bC(g)\sim b.$ Therefore, $b=cv$ for some unit $v\in D$ and $cf=bg=vcg$. Consequently, $f=vg$ (since $c\neq 0$), hence $f$ and $g$ are associates in $D[x]$. ($\implies$) Clear, since if $f=gu$ for some $u\in D[x]\subseteq F[x]$, then $f$ and $g$ are associates in $F[x]$. Primitive $f$ is irreducible in $D[x]$ iff $f$ is irreducible in $F[x]$ Let $D$ be a UFD with quotient field $F$ and $f$ a primitive polynomial of positive degree in $D[x]$. Then $f$ is irreducible in $D[x]$ if and only if $f$ is irreducible in $F[x]$. Proof ($\implies$) Suppose $f$ is irreducible in $D[x]$ and $f=gh$ with $g,h\in F[x]$ and $\deg g\geq 1$, $\deg h\geq 1$. Then $g=\sum_{i=0}^n(a_i/b_i)x^i$ and $h=\sum_{j=0}^m(c_j/d_j)x^j$ with $a_i, b_i, c_j, d_j\in D$ and $b_i\neq 0$, $d_j\neq 0$. Let $b=b_0b_1\dots b_n$ and for each $i$ let $\displaystyle b_i^*=b_0b_1\dots b_{i-1}b_{i+1}\dots b_n.$ If $g_1=\sum_{i=0}^n a_ib_i^* x^i\in D[x]$ (clear denominators of $g$ by multiplying by product of denominators), then $g_1=ag_2$ with $a=C(g_1)$, $g_2\in D[x]$ and $g_2$ primitive. Verify that $g=(1_D/b)g_1=(a/b)g_2$ and $\deg g=\deg g_2$. Similarly $h=(c/d)h_2$ with $c,d\in D$, $h_2\in D[x]$, $h_2$ primitive and $\deg h=\deg h_2$. Consequently, $f=gh=(a/b)(c/d)g_2h_2$, hence $bdf=acg_2h_2$. Since $f$ is primitive by hypothesis and $g_2h_2$ is primitive by Gauss Lemma, $\displaystyle bd\sim bdC(f)\sim C(bdf)=C(acg_2h_2)\sim acC(g_2h_2)\sim ac.$ This means $bd$ and $ac$ are associates in $D$. Thus $ubd=ac$ for some unit $u\in D$. So $f=ug_2h_2$, hence $f$ and $g_2h_2$ are associates in $D[x]$. Consequently $f$ is reducible in $D[x]$ (since $f=ug_2h_2$), which is a contradiction. Therefore, $f$ is irreducible in $F[x]$. ($\impliedby$) Conversely if $f$ is irreducible in $F[x]$ and $f=gh$ with $g,h\in D[x]$, then one of $g$, $h$ (say $g$) is a unit in $F[x]$ and thus a (nonzero) constant. Thus $C(f)=gC(h)$. Since $f$ is primitive, $g$ must be a unit in $D$ and hence in $D[x]$. Thus $f$ is irreducible in $D[x]$. Posted in math | Tagged , | Leave a comment Animation: Linear Algebra Abstract Vector Spaces ​向量空间 Eigenvalues & Eigenvectors (valeurs propres et vecteurs propres) 特征值/特征向量 [ Note: “Eigen-” is German for Characteristic 特征.] The Essence of Determinant (*): (行列式) (*) Determinant was invented by the ancient Chinese Algebraists 李冶 / 朱世杰 /秦九韶 in 13th century (金 / 南宋 / 元) in《天元术》.The Japanese “和算” mathematician 关孝和 spread it further to Europe before the German mathematician Leibniz named it the “Determinant” in 18th century. [NOTE] 金庸 武侠小说 《神雕侠女》里 元朝初年的 黄蓉 破解 大理国王妃 瑛姑 苦思不解的 “行列式”, 大概是求 eigenvalues & eigenvectors ? 🙂 View original post Posted in math | Leave a comment Cours Diagonalisation 1/2. Cours Maths spé Maths Spéciales (2nd Year French Engineering University Math ”Linear Algebra”): The French way of treating Matrices is very general and abstract. Advantage is it studies Matrices at a theoretical high-level, disadvantage is it ignores on applications. This young French prof explained in 30 mins at great length of what is simply the Characteristic Polynomial Equation:latex boxed {det (A – lambda.I) = 0 }&fg=aa0000&s=3$Compare it with the more practical (but less theoretical – 知其然而不知其所以然) American teaching below from the famous MIT Prof Gilbert Strang for the same Diagonalization of Matrices: View original post Posted in math | Leave a comment Free Math Notes by AMS Just learnt from Professor Terence Tao’s blog that there is a new series of free math notes by the American Mathematical Society: http://www.ams.org/open-math-notes. Many of the notes there are of exceptionally high quality (check out “A singular mathematical promenade”, by Étienne Ghys). Welcome to AMS Open Math Notes, a repository of freely downloadable mathematical works in progress hosted by the American Mathematical Society as a service to researchers, teachers and students. These draft works include course notes, textbooks, and research expositions in progress. They have not been published elsewhere, and, as works in progress, are subject to significant revision. Visitors are encouraged to download and use these materials as teaching and research aids, and to send constructive comments and suggestions to the authors. Posted in math | Tagged | Leave a comment Non-trivial submodules of direct sum of simple modules Suppose $M_1$ and $M_2$ are two non-isomorphic simple, nonzero $R$-modules. Determine all non-trivial submodules of $M_1\oplus M_2$. Let $N$ be a non-trivial submodule of $M_1\oplus M_2$. Note that $\displaystyle \{0\}\subset M_1\subset M_1\oplus M_2$ is a composition series. By Jordan-Holder theorem, all composition series are equivalent and have the same length. Hence $\displaystyle \{0\}\subset N\subset M_1\oplus M_2$ must be a composition series too. Thus $N\cong M_1$ or $M_2$. In particular $N$ is simple. Let $\pi_1: M_1\oplus M_2\to M_1$ and $\pi_2:M_1\oplus M_2\to M_2$ be the canonical projections. Note that $\pi_1(N)$ is a submodule of $M_1$, so $\pi_1(N)\cong 0$ or $M_1$. Similarly, $\pi_2(N)\cong 0$ or $M_2$. By Schur’s Lemma $\pi_1|_N: N\to \pi_1(N)$ and $\pi_2|_N: N\to\pi_2(N)$ are either 0 or isomorphisms. They cannot be both zero since $N$ is non-zero. They cannot be both isomorphisms either, as that would imply $M_1\cong\pi_1(N)\cong\pi_2(N)\cong M_2$. Hence, exactly one of $\pi_1$, $\pi_2$ are zero. So $N=M_1\oplus\{0\}$ or $\{0\}\oplus M_2$. Posted in math | Tagged | Leave a comment Recommended Educational Toys Does your child complain that science is “boring”? This may be because science is often taught in a boring manner. The solution may be to supplement teaching with hands-on experiments that develop the inner curiosity of the child. For parents looking to buy an educational toy for their child, here are two recommendations: For its price, it is one of the most affordable microscopes around. Suitable as a starter microscope for children interested in doing experiments. Suitable for upper primary (Grade 5/6) onwards. For more serious/experienced students, they can consider AmScope B120C-E1 Siedentopf Binocular Compound Microscope, 40X-2500X Magnification, LED Illumination, Abbe Condenser, Two-Layer Mechanical Stage, 1.3MP Camera and Software Windows XP/Vista/7/8/10 which is probably even better than the microscope in your secondary school / junior college. It can be connected to the computer for deeper analysis. Snap Circuits Jr. SC-100 Electronics Discovery Kit Electricity is one of the greatest inventions in the past century, and also a key component of the science syllabus from primary all the way to university. Learn more about circuits in this amazing toy. Suitable for lower primary onwards (Grade 1-3). Posted in math | Tagged , | 1 Comment Kids with tuition fare worse? Those who read the news, either online or in print, would probably have seen this article: “Kids with tuition fare worse”. In the article, it is claimed that: “In fact, children who received tuition actually scored about 0.256 standard deviations lower on their tests than those who did not (standard deviation is a measure of how spread out test scores are from the average).” The headline is actually quite misleading, causing people to think that tuition causes worse performance. One needs to read the final part of the article: “The first is that students who receive tuition choose to receive it precisely because they are not doing well in school. In other words, weak performance may be what is driving students to enrol for tuition.” The correct way to measure the effect of tuition is via a “before and after” experiment. Scores of students before and after enrolling in tuition should be compared to truly see if tuition has any effect. Many tuition centers are already doing this, it is not a rocket science experiment. Without the “before and after” comparison, the research is meaningless. It is like saying, “People who see a medical doctor frequently have poorer health.”, it is true, but obviously one cannot conclude that medical doctors cause poor health! Lastly, the research is analysing PISA data (Programme for International Student Assessment). Clearly, there is no tuition centre tutoring PISA, which is significantly different from the ordinary curriculum (I was a PISA grader). As tuition is highly specialized, it is true that tuition can have close to zero effect on PISA scores. It is like PSLE / O Level Math tuition has close to no effect on Math Olympiad scores; even if it is both “Math”, it is possible to score full marks in PSLE / O Level Math but zero marks in Olympiad Math! Posted in math | Tagged , | 1 Comment Real Life Tortoise vs. Hare Quite motivational indeed! Posted in math | Tagged | 1 Comment Commutator subgroup G’ is the unique smallest normal subgroup N such that G/N is abelian. Commutator subgroup $G'$ is the unique smallest normal subgroup $N$ such that $G/N$ is abelian. If $G$ is a group, then $G'$ is a normal subgroup of $G$ and $G/G'$ is abelian. If $N$ is a normal subgroup of $G$, then $G/N$ is abelian iff $N$ contains $G'$. Proof Let $f:G\to G$ be any automorphism. Then $\displaystyle f(aba^{-1}b^{-1})=f(a)f(b)f(a)^{-1}f(b)^{-1}\in G'.$ It follows that $f(G')\leq G'$. In particular, if $f$ is the automorphism given by conjugation by $a\in G$, then $aG'a^{-1}=f(G')\leq G'$, so $G'\unlhd G$. Since $(ab)(ba)^{-1}=aba^{-1}b^{-1}\in G'$, $abG'=baG'$ and hence $G/G'$ is abelian. ($\implies$) If $G/N$ is abelian, then $abN=baN$ for all $a,b\in G$. Hence $ab(ba)^{-1}=aba^{-1}b^{-1}\in N$. Therefore, $N$ contains all commutators and $G'\leq N$. ($\impliedby$) If $G'\subseteq N$, then $ab(ba)^{-1}=aba^{-1}b^{-1}\in G'\subseteq N$. Thus $abN=baN$ for all $a,b\in G$. Hence $G/N$ is abelian. Posted in math | Tagged | Leave a comment Ascending Central Series and Nilpotent Groups Ascending Central Series of $G$ Let $G$ be a group. The center $C(G)$ of $G$ is a normal subgroup. Let $C_2(G)$ be the inverse image of $C(G/C(G))$ under the canonical projection $G\to G/C(G)$. By Correspondence Theorem, $C_2(G)$ is normal in $G$ and contains $C(G)$. Continue this process by defining inductively: $C_1(G)=C(G)$ and $C_i(G)$ is the inverse image of $C(G/C_{i-1}(G))$ under the canonical projection $G\to G/C_{i-1}(G)$. Thus we obtain a sequence of normal subgroups of $G$, called the ascending central series of $G$: $\displaystyle \langle e\rangle Nilpotent Group A group $G$ is nilpotent if $C_n(G)=G$ for some $n$. Abelian Group is Nilpotent Every abelian group $G$ is nilpotent since $G=C(G)=C_1(G)$. Every finite $p$-group is nilpotent (Proof) $G$ and all its nontrivial quotients are $p$-groups, and therefore have non-trivial centers. Hence if $G\neq C_i(G)$, then $G/C_i(G)$ is a $p$-group, and $C(G/C_i(G))$ is non-trivial. Thus $C_{i+1}(G)$, the inverse image of $C(G/C_i(G))$ under $\pi:G\to G/C_i(G)$, strictly contains $C_i(G)$. Since $G$ is finite, $C_n(G)$ must be $G$ for some $n$. Posted in math | Tagged | Leave a comment Singaporean Student Wins US$250K Scholarship

Congratulations to Ms See for the win. The amount of the scholarship is no joke, it is the price of a HDB flat (a home in Singapore).

If you are interested to take part, here are the details: The Breakthrough Junior Challenge is an annual competition for students, ages 13-18, to share their passion for math and science with the world! In partnership with the Khan Academy, each student submits a video that explains a challenging and important concept or theory in mathematics, life sciences, or physics. The winner receives a $250,000 college scholarship. The winning student’s teacher and school also benefit:$50,000 for the teacher and a state-of-the-art 100,000 science lab for the school. Learn more at https://breakthroughjuniorchallenge.org. AsianScientist (Dec. 6, 2016) – Ms Deanna See, a 17-year-old studying at Raffles Institution in Singapore, was one of two students to win the Breakthrough Junior Challenge, a global science video competition that is part of the 2017 Breakthrough Prizes. See’s fun and quirky five-minute video, titled ‘Superbugs! And Our Race Against Resistance,’ explains the phenomenon of antibiotic resistance with the use of Lego figurines. She takes the viewer through the evolution and genetics behind how superbugs amass their formidable defenses against various antibiotics. Watch See’s video below: Read more from Asian Scientist Magazine at: http://www.asianscientist.com/2016/12/topnews/deanna-see-breakthrough-junior-challenge-singapore/ The other finalist, Antonella Masini from Peru Posted in math | Tagged | Leave a comment Existence of Splitting Field with degree less than n! If $K$ is a field and $f\in K[X]$ has degree $n\geq 1$, then there exists a splitting field $F$ of $f$ with $[F:K]\leq n!$. Proof: We use induction on $n=\deg f$. Base case: If $n=1$, or if $f$ splits over $K$, then $F=K$ is a splitting field with $[F:K]=1\leq 1!$. Induction Hypothesis: Assume the statement is true for degree $n-1$, where $n>1$. If $n=\deg f>1$ and $f$ does not split over $K$, let $g\in K[X]$ be an irreducible factor of $f$ with $\deg g>1$. Let $u$ be a root of $g$, then $\displaystyle [K(u):K]=\deg g>1.$ Write $f=(x-u)h$ with $h\in K(u)[X]$ of degree $n-1$. By induction hypothesis, there exists a splitting field $F$ of $h$ over $K(u)$ with $[F:K(u)]\leq(n-1)!$. That is, $h=u_0(x-u_1)\dots(x-u_{n-1})$ with $u_i\in F$ and $F=K(u)(u_1,\dots,u_{n-1})=K(u,u_1,\dots,u_{n-1})$. Thus $f=u_0(x-u)(x-u_1)\dots(x-u_{n-1})$, so $f$ splits over $F$. This shows $F$ is a splitting field of $f$ over $K$ of dimension \begin{aligned} [F:K]&=[F:K(u)][K(u):K]\\ &\leq (n-1)!(\deg g)\\ &\leq n! \end{aligned} Posted in math | Tagged | Leave a comment Fate: A Hebrew Folktale A Hebrew Folktale King Solomon’s servant came breathlessly into the court, “Please! Let me borrow your fastest horse!” he said to the King. “I must be in a town ten miles south of here by nightfall!” “Why?” asked King Solomon. “Because,” said his shuddering servant, “I just met Death in the garden! Death looked me in the face! I know for certain I’m to be taken and I don’t want to be around when Death comes to claim me!” “Very well,” said King Solomon. “My fastest horse has hoofs like wings. TAKE HIM.” Then Solomon walked into the garden. He saw Death sitting there with a perplexed look on its face. “What’s wrong?” asked King Solomon. Death replied, “Tonight I’m supposed to claim the life of your servant whom I just now saw in your garden. But I’m supposed to claim him in a town ten miles south of here! Unless he had a horse with hooves like wings, I don’t see how he could get there by nightfall . . .” Posted in math | Tagged | 1 Comment Transitivity of Algebraic Extensions Let $K\subseteq E\subseteq F$ be a tower of fields. If $F/E$ and $E/K$ are algebraic, then $F/K$ is algebraic. Proof: (Hungerford pg 237, reworded) Let $u\in F$. Since $u$ is algebraic over $E$, there exists some $b_i\in E$ ($b_n\neq 0$) such that $\displaystyle b_0+b_1u+\dots+b_nu^n=0.$ Let $L=K(b_0,\dots,b_n)$, then $u$ is algebraic over $L$. Hence $L(u)/L$ is finite. Note that $L/K$ is finitely generated and algebraic, since each $b_i\in E$ is algebraic over $K$. Thus $L/K$ is finite. Thus by Tower Law, $L(u)/K$ is finite, thus algebraic. Hence $u\in L(u)$ is algebraic over $K$. Since $u$ was arbitrary, $F$ is algebraic over $K$. Posted in math | Tagged | Leave a comment Counterexamples to Normal Extension Let $K\subseteq L\subseteq M$ be a tower of fields. Q1) If M/K is a normal extension, is L/K a normal extension? False. Let $M$ be the algebraic closure of $K=\mathbb{Q}$. Let $L=\mathbb{Q}(\sqrt[3]{2})$. Then $M$ is certainly a normal extension of $\mathbb{Q}$ since every irreducible polynomial in $\mathbb{Q}[X]$ that has one root in $M$ has all of its roots in $M$. However consider $X^3-2\in\mathbb{Q}[X]$. It has one root $(\sqrt[3]{2})$ in $L$, but the other two complex roots are not in $L$. Thus $L/K$ is not a normal extension. Q2) If M/L and L/K are both normal extensions, is M/K a normal extension? (i.e. is normal extension transitive?) False. Let $L=\mathbb{Q}(\sqrt 2)$, $K=\mathbb{Q}$. Then $L/K$ is normal since $L$ is the splitting field of $X^2-2$ over $\mathbb{Q}$. Let $M=\mathbb{Q}(\sqrt 2,\sqrt[4]{2})$. Then $M/L$ is normal since $M$ is the splitting field of $X^2-\sqrt 2$ over $L$. However, $M/K$ is not normal. The polynomial $X^4-2$ has a root in $M$ (namely $\pm\sqrt[4]{2}$) but the other two complex roots are not in $M$. Posted in math | Tagged | Leave a comment Feng Tianwei wins World No. 1 Ding Ning despite being dropped from Singapore Team Congratulations to Feng Tianwei for the epic win over World No. 1 Ding Ning in the Chinese Super League, despite being recently axed from the Singapore national team (for unknown reasons). The Chinese Super League, though not as famous as the Olympics, is obviously much harder and tougher to win than the Olympic Games. The main reason being that the entire platoon of All-Star China Team is playing there, while in the Olympics China is restricted to sending 3 men and 3 women. So, once again congratulations, and Feng’s perseverance and fighting spirit is very motivating to all. Posted in math | | 1 Comment Top PSLE Score The top PSLE Score for this year seems to be 286, from RGPS (Raffles Girls Primary School). Close runners-ups are 283, from NYPS (Nanyang Primary School) and Nanhua. Source is from https://www.kiasuparents.com/kiasu/t-scores/, which is self-reported by parents. Casting aside the “troll scores” of 299 or 300 which are not believable, this seems to be the most accurate top PSLE score available, since the mainstream media are not allowed to report them. Congratulations for those who have done well. And for those who have not, do not be discouraged as there is still a long road ahead, and there will be many opportunities to prove yourself. For those considering tuition, check out StarTutor, which is highly recommended by us. Tutors are screened for their educational qualifications and matched accordingly, with zero administrative fees. From experience, O Level is a completely new ball game from PSLE. It is possible for relatively weak students in PSLE to do very well in O Levels, and also vice versa; for quite strong PSLE students to do poorly in O Levels. It is a new beginning for students. Posted in math | Tagged , | Leave a comment Motivational Quote by Sport Psychologist “The reality is that if your dream is to accomplish something awesome, it’s not going to be easy. If it were easy, everyone would be doing it. People who go for greatness are going to get knocked down a lot. They’ll have difficult times. They’ll struggle with doubt and uncertainty. People around them will question the wisdom of their quest. The issue is not whether you’ll fail, because you will. It’s whether you’ll get back up and keep going. It’s whether you can sustain your self-confidence and your belief in yourself and keep bouncing back. Failure is only final when you stop striving.” – Bob Rotella Posted in math | Tagged | Leave a comment Proof of Equivalent Conditions for Split Exact Sequence Attached is a proof of the equivalent conditions for a Split Exact Sequence, based on the nice proof in Hungerford using the Short Five Lemma. Very neat proof. Split Exact Sequence Proof Posted in math | Tagged , | Leave a comment Recent Interview of Shing-Tung Yau (in Chinese) Excellent interview of S.T. Yau, Fields Medalist. One mischievous student tried to ask a trick question that is a variant of the Missing Dollar Problem. The interviewer is Sa Beining, who is a famous celebrity in China. Not much mathematical content though, since it is aimed at the general audience. Nevertheless, it is inspirational, especially for Chinese youth. Posted in math | Tagged , | Leave a comment The Reason Why Singaporean Students are Top in Maths (PISA) Quite interesting analysis on how and why Singapore topped the ranking for PISA in Math/Science. One possible reason is the difficulty of PSLE trains students to solve tricky and difficult (for that level) math questions. It is well known that PSLE questions are more difficult (for students of their respective levels) than O Level E Maths questions. The peak difficulty in Singapore Maths Syllabus are at PSLE and then at H2 A Level Maths. Overall, on average, the average Singaporean student is quite well-prepared in maths. Since PISA is measuring the average capability of students (rather than the top echelon of students), it is no surprise that Singapore would score rather highly in this aspect. Note that China (Mainland) seems to be missing from the study. If China (especially if restricted to cities like Shanghai and Beijing) are included, they would be a strong contender for No. 1 position too, since it is well known that China Math is just as difficult, or even more so, especially at high school/senior high school level. Source (in Chinese): http://mp.weixin.qq.com/s/lQx4kmqMwNsZEUlXQbRIzw 最近一个报告，说新加坡学中小学生的数理能力，在全球64个国家中，排名第一！ （阅读提示，本文有点长，没耐心看废话的，直接拉到最后看一下就好。） 引起不少网友的讨论。感觉上，最会算的，不应该是华人尤其是中国人吗？好吧，作为华人咱谦虚一点，犹太人俄罗斯人甚至美国人印度人，都有可能第一啊，数完一个巴掌也轮不到新加坡哇，又没解歌德巴赫猜想，也没研究天体物理出啥成果，出产的程序员不够内需，还要进口外援，凭什么？ 一般江湖传说，中国中小学生到了国外，数学都能甩当地同级学生几条街。 我们来看看新加坡的数学题 问：8个1元硬币大概有多重？备选答案 6克； 60克； 600克； 6千克。 或许有人会第一反应，这是奥数题。毕竟要考的点很多，生活常识、逻辑能力……，一般学生可不会碰到这样的考题。 事实上，它是去年新加坡小学离校会考数学题中的一道。 没有答对的读者们，你们小学能毕业吗？（幸灾乐祸脸） 有些孩子不懂怎么做，胡乱猜了一个答案，有些家长（新加坡眼觉得是恼羞成怒）就投诉考题超纲。 新加坡教育部的回应有理有据，考题可是根据教学大纲准备的，到底是大纲的哪一条，都能翻给你看。 说到新加坡的教育大纲，那也是杠杠的。 美国居然也采用新加坡的数理教材！足见根据新加坡大纲编写的教材，相当有章法。 (Translation: America is also using Singapore Math material to teach their students!) 新加坡小学生数理能力第一咋来的 在最近公布的《国际数学与科学趋势研究报告》。 一个国际教育组织，每四年都会随机抽取小学四年级和中学二年级的学生，做国际数学与科学趋势研究（简称TIMSS）。 去年，新加坡各中小学里，共有6500名小四学生和6100中二学生接受了调查，要跟64个国家和城市的学生一比高低。为了拿到更直观和准确的数据，学生们被分为四类：基本、中水平、高水平，优等，进行考察。 先来围观，这份调查是从那些方面着手的？ 国际教育成就评估协会以试卷评估学生这些方面：（其实就是参加考试） • 对知识的掌握能力； • 应用能力； • 推理能力。 另外还有学生们对学习的态度等等。 考试的内容在这里~ 看成绩~~以中二学生的成绩为例（有兴趣的网友，可以点击文章底部“阅读原文”，查看全报告详情） 科学方面，新加坡597分。第二名是日本，571分，前后相差26分，差距还是不小，第三名是中国台北569分。 点击看大图 数学方面，新加坡621分，第二名韩国606分；第三名中国台北599分；第四名中国香港594分；第五名日本586分。 点击看大图 新加坡人的数理能力，到底强不强，我们听听在籍学生、资深数学补习老师、家有才女的、中国留学生，各方的意见~~ 新加坡的资深数学补习老师，孙老师，在中国和新加坡分别有19年和8年的教学经验，教过中国的高中和大学，也在新加坡教过O level、PSLE考生，涉猎高等数学，并编写O Level 数学中英文教材，主要适用中国留学生，被私立学校广泛使用。 这样的排名可以理解。毕竟本地的孩子和家长都非常努力，想不拿第一都难。 1）本地数学题的特点：不出偏怪题，会让那些努力但不聪明的孩子，也有能力取得好成绩，不会失去学习兴趣。 2）新加坡的数学在某种程度上要比其他国家难。尤其是小五小六的数学word problems（相当于中国的应用题），比中国更难，因为不能用方程解，本地学生从小二就开始画model。 只有那些很聪明的孩子才能应付。（这句话真是说到了编辑部某位学妈的心坎里，回想当年陪娃儿读书，明明可以方程解，设个X，一元一次方程，不是很容易搞定吗，硬要从头学起画model，真是郁闷死了），所以本地有不少小五小六学生，会补习数学。毕竟从小五到高中毕业的A水准，要想拿A的话，还是挺有难度的。 传说中的Model，你看得懂吗？ 3）至于中学数学，要比中国学得广，但没有中国深。本地更注重解题过程。数学的应用性也比中国强。它的O水准数学考试题目，时间比中国的中考长，难度比中国的也要大些。大概做个比较的话，本地的O水准高等数学一科，相当于中国高三的数学了（编者按：相对于中国高考考三天大概五六门，新加坡的O水准可是要考至少一个月的，而且像数学，也分好多子科目，单看所有科目的编号都是4位数的，就知道名堂特别多。） 今年O水准数学考试的部分列表 4）本地的A水平数学题目，像写论文。所以学生的推理能力必须很强，才能解答。 5）相比中国的数学注重计算，速度以及一题多解，所以中国孩子的基础非常好。如果在中国有经过4年的小学数学训练，再来新加坡的话，基本就不要补习了。尤其那些在中国学霸级的学生来新后，一定还是学霸。 Posted in math | Tagged , | 2 Comments Who cares about topology? (inscribed rectangle problem) Excellent video for the curious minds! Who cares about Topology such as Torus (aka donut) or Mobius Strip ? They can be used to prove difficult math such as the unsolved problem “Inscribed square/rectangle inside any closed loop”. To understand the Topology on Loops, please view the lecture here : Homotopyand the fundamental group of surface. View original post Posted in math | Leave a comment Conditions for S^-1I=S^-1R (Ring of quotients) Conditions for $S^{-1}I=S^{-1}R$: Let $S$ be a multiplicative subset of a commutative ring $R$ with identity and let $I$ be an ideal of $R$. Then $S^{-1}I=S^{-1}R$ if and only if $S\cap I\neq\varnothing$. Proof (H pg 146) $(\implies)$ Assume $S^{-1}I=S^{-1}R$. Consider the ring homomorphism $\displaystyle \phi_S: R\to S^{-1}R$ given by $r\mapsto rs/s$ (for any $s\in S$). Then $\phi_S^{-1}(S^{-1}I)=R$ hence $\phi_S(1_R)=a/s$ for some $a\in I$, $s\in S$. Since $\phi_S(1_R)=1_Rs/s$, we have $s_1(s^2-as)=0$ for some $s_1\in S$, i.e.\ $s^2s_1=ass_1$. But $s^2s_1\in S$ and $ass_1\in I$ imply $S\cap I\neq\varnothing$. $(\impliedby)$ If $s\in S\cap I$, then $1_{S^{-1}R}=s/s\in S^{-1}I$. Note that for any $r/s\in S^{-1}R$, $\displaystyle (\frac{r}{s})(\frac{s}{s})=\frac{rs}{s^2}=\frac{r}{s}\in S^{-1}I$ since $rs\in I$ and $s^2\in S$. Thus $S^{-1}I=S^{-1}R$. Posted in math | Tagged | Leave a comment Local Ring Equivalent Conditions If $R$ is a commutative ring with 1 then the following conditions are equivalent. (i) $R$ is a local ring, that is, a commutative ring with 1 which has a unique maximal ideal. (ii) All nonunits of $R$ are contained in some ideal $M\neq R$. (iii) The nonunits of $R$ form an ideal. Proof (H pg 147) (i)$\implies$(ii): If $a\in R$ is a nonunit, then $(a)\neq R$ since $1\notin (a)$. Therefore $(a)$ (and hence $a$) is contained in the unique maximal ideal $M$ of $R$, since $M$ must contain every ideal of $R$ (except $R$ itself). (ii)$\implies$(iii): Let $S$ be the set of all nonunits of $R$. We have $S\subseteq M\neq R$. Let $x\in M$. Since $M\neq R$, $x$ cannot be a unit. So $x\in S$. Thus $M\subseteq S$. Hence $S=M$, which is an ideal. (iii)$\implies$(i): Assume $S$, the set of nonunits, form an ideal. Let $I\neq R$ be a maximal ideal. Let $a\in I$, then $a$ cannot be a unit so $a\in S$. Thus $I\subseteq S\neq R$. By maximality $S=I$ and this shows $S$ is the unique maximal ideal. Posted in math | Tagged | Leave a comment Lp Interpolation It turns out that there are two types of Lp interpolation: One is called “Lyapunov’s inequality” which is addressed in this previous blog post. The other one is called Littlewood’s inequality: If $f\in L^p\cap L^q$, then $f\in L^r$ for any intermediate $p. The proof is here: Case 1) $q<\infty$. We have $\frac{1}{q}<\frac{1}{r}<\frac{1}{p}$. There exists $0<\lambda<1$ such that $\frac{1}{r}=\frac{\lambda}{p}+\frac{1-\lambda}{q}$. \begin{aligned} \|f\|_r^r&=\int |f|^r\\ &=\int |f|^{\lambda r}|f|^{(1-\lambda)r}\\ &\leq\left(\int |f|^{\lambda r(\frac{p}{\lambda r})}\right)^{\lambda r/p}\left(\int |f|^{(1-\lambda)r(\frac{q}{(1-\lambda)r})}\right)^{r(1-\lambda)/q}\\ &=\left(\int |f|^p\right)^{\lambda r/p}\left(\int |f|^q\right)^{r(1-\lambda)/q}\\ &=\|f\|_p^{\lambda r}\|f\|_q^{r(1-\lambda)}. \end{aligned} Thus $\|f\|_r\leq\|f\|_p^\lambda\|f\|_q^{1-\lambda}$. Case 2) $q=\infty$. We have $0<\frac{1}{r}<\frac{1}{p}$. There exists $0<\lambda<1$ with $\frac{1}{r}=\frac{\lambda}{p}$. Then \begin{aligned} \|f\|_r^r&=\||f|^{\lambda r}|f|^{(1-\lambda)r}\|_1\\ &\leq\||f|^{\lambda r}\|_{\frac{p}{\lambda r}}\||f|^{(1-\lambda)r}\|_\infty\\ &=\left(\int |f|^p\right)^{\lambda r/p}\|f\|_\infty^{(1-\lambda)r}\\ &=\|f\|_p^{\lambda r}\|f\|_\infty^{(1-\lambda)r}. \end{aligned} Hence $\|f\|_r\leq\|f\|_p^\lambda\|f\|_\infty^{1-\lambda}$. Posted in math | Tagged | Leave a comment Homotopy NJ Wildberger AlgTop24: The fundamental group Homotopy 同伦: When playing skip rope, the 2 ends of the rope are held by 2 persons while a 3rd person jumping over the “swings of rope” – these swings at any instant arehomotopic. View original post Posted in math | Leave a comment Generalized Riemann-Stieltjes Integral The generalized Riemann-Stieltjes integral $\int_a^b f\,d\phi$ is a number $\gamma$ such that: for every $\epsilon>0$ there exists a partition $P_\epsilon$ of $[a,b]$ such that if $\dot{P}=(P,\xi)$, $P=\{a=x_0, $\xi=\{\xi_i: i=1,\dots, n\}$ with $\xi_i\in[x_{i-1},x_i]$ is a tagged partition of $[a,b]$ such that $P$ is a refinement of $P_\epsilon$, then $\displaystyle |S(\dot{P},f,\phi)-\gamma|=|\sum_{i=1}^n f(\xi_i)(\phi(x_i)-\phi(x_{i-1}))-\gamma|<\epsilon.$ We will write $\displaystyle \lim_{P\to 0}S(\dot{P},f,\phi)=\lim_{P\to 0}\sum_{i=1}^nf(\xi_i)(\phi(x_i)-\phi(x_{i-1}))=\gamma$ and $f\in \mathcal{R}_\phi[a,b]$. Posted in math | Tagged | Leave a comment Matrix A.X = 0 1) Row reduction, row-echelon form and reduced row-echelon form 2) Rank of Matrix = Rank (A) View original post Posted in math | Leave a comment Normalizer of Normalizer of Sylow p-subgroup The normalizer of a Sylow p-subgroup is “self-normalizing”, i.e. its normalizer is itself. Something that is quite cool. If $P$ is a Sylow $p$-subgroup of a finite group $G$, then $N_G(N_G(P))=N_G(P)$. Proof (Adapted from Hungerford pg 95) Let $N=N_G(P)$. Let $x\in N_G(N)$, so that $xNx^{-1}=N$. Then $xPx^{-1}$ is a Sylow $p$-subgroup of $N\leq G$. Since $P$ is normal in $N$, $P$ is the only Sylow $p$-subgroup of $N$. Therefore $xPx^{-1}=P$. This implies $x\in N$. We have proved $N_G(N_G(P))\subseteq N_G(P)$. Let $y\in N_G(P)$ Then certainly $yN_G(P)y^{-1}=N_G(P)$, so that $y\in N_G(N_G(P))$. Thus $N_G(P)\subseteq N_G(N_G(P))$. Posted in math | Tagged | Leave a comment Index of smallest prime dividinglatex |G|\$ implies Normal Subgroup

I have previously proved this at: Advanced Method for Proving Normal Subgroup. This is a neater, slightly shorter proof of the same theorem.

Index of smallest prime dividing $|G|$ implies Normal Subgroup
If $H$ is a subgroup of a finite group $G$ of index $p$, where $p$ is the smallest prime dividing the order of $G$, then $H$ is normal in $G$.

Proof:
(Hungerford pg 91)

Let $G$ act on the set $G/H$ (left cosets of $H$ in $G$) by left translation.

This induces a homomorphism $\sigma: G\to S_{G/H}\cong S_p$, where $\sigma_g(xH)=gxH$. Let $g\in\ker\sigma$. Then $gxH=xH$ for all $xH\in G/H$. In particular, when $x=1$, $gH=H$ which implies $g\in H$. So we have $\ker\sigma\subseteq H$.

Let $K=\ker\sigma$. By First Isomorphism Theorem, $G/K\cong\text{Im}\,\sigma\leq S_p$. Hence $|G/K|$ divides $|S_p|=p!$ But every divisor of $|G/K|=[G:K]$ must divide $|G|=|K|[G:K]$. Since no number smaller than $p$ (except 1) can divide $|G|$, we must have $|G/K|=p$ or $1$. However $\displaystyle |G/K|=[G:K]=[G:H][H:K]=p[H:K]\geq p.$

Therefore $|G/K|=p$ and $[H:K]=1$, hence $H=K$. But $K=\ker\sigma$ is normal in $G$.

Normal Extension

An algebraic field extension $L/K$ is said to be normal if $L$ is the splitting field of a family of polynomials in $K[X]$.

Equivalent Properties
The normality of $L/K$ is equivalent to either of the following properties. Let $K^a$ be an algebraic closure of $K$ containing $L$.

1) Every embedding $\sigma$ of $L$ in $K^a$ that restricts to the identity on $K$, satisfies $\sigma(L)=L$. In other words, $\sigma$, is an automorphism of $L$ over $K$.
2) Every irreducible polynomial in $K[X]$ that has one root in $L$, has all of its roots in $L$, that is, it decomposes into linear factors in $L[X]$. (One says that the polynomial splits in $L$.)

Kiasuparents PSLE

Basically to summarize the article above, the co-founder of Kiasuparents’ son scored a respectable 4As and 229 T-score for PSLE. However, as their set target was 250, he did not get the Nintendo DS that was part of the deal for achieving the target of 250. Probably the Nintendo is to play the most recently released Pokemon Sun/Moon. Poor kid! I remember that my highlight of finishing PSLE was to play Pokemon (close to 20 years ago). I still remembered I was playing the Blue version, starting as Bulbasaur.

PSLE can be highly unpredictable (variance of 20-30 marks from usual expected mark is common and expected). This is particularly due to language exams, composition, and also the famous rigid marking scheme of PSLE science, where all the “keywords” must be mentioned in order to get the mark. Mathematics is the more reliable subject here as it is more objective, so try to score as high as possible in it.

Hence DSA becomes increasingly important as a backup plan to act as insurance in the event that something goes wrong in the PSLE. Check out some DSA/GAT/HAST posts here. It is always good to have a “Plan B”.

Also, if you suspect that the child’s school teacher is not that excellent in teaching, e.g. don’t know/emphasize the “keywords” which are necessary to get any marks at all in PSLE science, you may consider engaging a tutor as soon as possible. Check out the most recommended tuition agency in Singapore.

A Journey from Undergrad Math in China to PhD Math in France

An autobiography of a Chinese PhD student  (France, University  Paris-11) in Number Theory and Algebraic  Geometry.
His journey from 武汉大学 4 years undergrad to Beijing, learn mostly from the French-educated Chinese Math professors (Ecole Normale  Superieure, Polytechniques).

“趁着一丝冲动记下一些经历，也趁着现在还能想起来，写写路途中遇到的人和碰到的一些书。”
http://blog.sina.cn/dpool/blog/s/blog_4ee63ce90102ea2r.html

Ecole Normale Supérieure (Paris) (E.N.S.): 巴黎高等师范大学 (Galois 的母校但他因参加法国革命被开除, 现在是法国/世界 Fields Medals 的摇篮, 出Bourbaki 学派的大师André Weil, Cartan, Dieudoné, …医学细菌发现者 Pasteur是排班上最后一名的劣等生)。E.N.S.训练未来的教授 (文, 数, 理), 每年只收全法国前50位精英学生, 培养成博士。后来演变成研究院, 出了不少 Nobel Prize (Science, Literature) 和Fields Medalists.

Paris University 11 = Paris Sud (Paris South). 欧洲最古老的巴黎大学(Sorbonne 索尔本), 继承十世纪阿拉伯人创办的大学制度 (Bachelor, Licencié, Baccalaureate, “Chair” of department…)。出科学家居里夫人 (Madame Curie)。现在有13分校, 其中第11分校是数学研究的重镇。

Ecole Polytechnique (aka X): 巴黎综合理工大学 (拿破仑建的工程军校, 出很多科学家, 数学家: Hertz, Ampere, Fourier, Cauchy, Poincaré, Louiville,…偏偏天才Galois 入学连续考2年Concours不及格, 学弟 Charles Hermite 入学考最后一名, 第二年又被踢出门)。

What is “Motif” (Motive) ?

Recommended Books:
1. Jean-Pierre Serre: 《Cours d’Arithmetique》

“Arithmetic”…

View original post 77 more words

Counting Homology Using Matrix

Column Space and Null Space of a matrix

Create your own Homology: (Important lecture in Applied Algebraic Topology)

View original post

Some Linear Algebra Theorems

Linear Algebra

Diagonalizable & Minimal Polynomial:
A matrix or linear map is diagonalizable over the field $F$ if and only if its minimal polynomial is a product of distinct linear factors over $F$.

Characteristic Polynomial:
Let $A$ be an $n\times n$ matrix. The characteristic polynomial of $A$, denoted by $p_A(t)$, is the polynomial defined by $\displaystyle p_A(t)=\det(tI-A).$

Cayley-Hamilton Theorem:
Every square matrix over a commutative ring satisfies its own characteristic equation:

If $A$ is an $n\times n$ matrix, $p(A)=0$ where $p(\lambda)=\det(\lambda I_n-A)$.

The Euler Characteristic χ

Euler Characteristicχis an invariant in Topology.

Interesting to note that a tree with no cycle (loop),χ =1

View original post

Free Abelian Groups & Free Vector Spaces

Algebraic Topology Foundation (Lecture 1 to 6). Very easy and helpful to undergrads!

Lecture 3: Modular Arithmetic (easy)https://youtu.be/5YEu-vB2dD8

Lecture 4: Addition and Free Vector Spaceshttps://youtu.be/9qL003DG7Og

View original post