Tag: Algebra

Structure Theorem for finitely generated (graded) modules over a PID

If R is a PID, then every finitely generated module M over R is isomorphic to a direct sum of cyclic R-modules. That is, there is a unique decreasing sequence of proper ideals (d_1)\supseteq(d_2)\supseteq\dots\supseteq(d_m) such that \displaystyle M\cong R^\beta\oplus\left(\bigoplus_{i=1}^m R/(d_i)\right) where d_i\in R, and \beta\in\mathbb{Z}.

Similarly, every graded module M over a graded PID R decomposes uniquely into the form \displaystyle M\cong\left(\bigoplus_{i=1}^n\Sigma^{\alpha_i}R\right)\oplus\left(\bigoplus_{j=1}^m\Sigma^{\gamma_j}R/(d_j)\right) where d_j\in R are homogenous elements such that (d_1)\supseteq(d_2)\supseteq\dots\supseteq(d_m), \alpha_i, \gamma_j\in\mathbb{Z}, and \Sigma^\alpha denotes an \alpha-shift upward in grading.

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Homogenous / Graded Ideal

Let A=\bigoplus_{i=0}^\infty A_i be a graded ring. An ideal I\subset A is homogenous (also called graded) if for every element x\in I, its homogenous components also belong to I.

An ideal in a graded ring is homogenous if and only if it is a graded submodule. The intersections of a homogenous ideal I with the A_i are called the homogenous parts of I. A homogenous ideal I is the direct sum of its homogenous parts, that is, \displaystyle I=\bigoplus_{i=0}^\infty (I\cap A_i).

Universal Property of Quotient Groups (Hungerford)

If f:G\to H is a homomorphism and N is a normal subgroup of G contained in the kernel of f, then f “factors through” the quotient G/N uniquely.Universal Property of Quotient

This can be used to prove the following proposition:
A chain map f_\bullet between chain complexes (A_\bullet, \partial_{A, \bullet}) and (B_\bullet, \partial_{B,\bullet}) induces homomorphisms between the homology groups of the two complexes.

Proof:
The relation \partial f=f\partial implies that f takes cycles to cycles since \partial\alpha=0 implies \partial(f\alpha)=f(\partial\alpha)=0. Also f takes boundaries to boundaries since f(\partial\beta)=\partial(f\beta). Hence f_\bullet induces a homomorphism (f_\bullet)_*: H_\bullet (A_\bullet)\to H_\bullet (B_\bullet), by universal property of quotient groups.

For \beta\in\text{Im} \partial_{A,n+1}, we have \pi_{B,n}f_n(\beta)=\text{Im}\partial_{B,n+1}. Therefore \text{Im}\partial_{A,n+1}\subseteq\ker(\pi_{B,n}\circ f_n).

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.

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.

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.

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.

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<s\ \text{and}\ p\nmid a_s.

Similarly there is a least integer t such that \displaystyle p\mid b_j\ \text{for}\ j<t\ \text{and}\ p\nmid b_t.

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].

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.

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.

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<C_1(G)<C_2(G)<\cdots.

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.

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}

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.

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.

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.

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)).

Index of smallest prime dividing $latex |G|$ 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.)

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).

Image and Preimage of Sylow p-subgroups under Epimorphism

Suppose G and H are p-groups, and \phi:G\to H is a surjective homomorphism.

Then for any Sylow p-subgroup P of G, \phi(P) is a Sylow p-subgroup of H.

Conversely, for any Sylow p-subgroup Q of H, Q=\phi(P) for some Sylow p-subgroup P of G.

Proof:

By the First Isomorphism Theorem, G/\ker\phi\cong\phi(G)=H. Write N=\ker\phi. Then \phi(P)=\{pN:p\in P\}=PN/N.

Since P is a Sylow p-subgroup of G, [G:P] is relatively prime to p. Thus, [G:PN]=[G:P]/[PN:P] is also relatively prime to p.

Then \displaystyle [H:\phi(P)]=[G/N:PN/N]=[G:PN] is also relatively prime to p. Since \phi(P)\cong P/\ker\phi|_P, \phi(P) is a p-group, so \phi(P) is a Sylow p-subgroup of H.

Part 2: Let Q be a Sylow p-subgroup of H\cong G/N. Then by Correspondence Theorem, Q\cong K/N for some subgroup K with N\subseteq K\subseteq G.

Then, [G:K]=[H:Q] is relatively prime to p, so K contains a Sylow p-subgroup P.

Consider P/N\cong\phi(P)\subseteq Q\cong K/N. By previous part, \phi(P) is a Sylow p-subgroup of H, so \phi(P)=Q.

Aut(G)=Aut(H)xAut(K), where H, K are characteristic subgroups of G with trivial intersection

Let G=HK, where H, K are characteristic subgroups of G with trivial intersection, i.e. H\cap K=\{e\}. Then, Aut(G)=Aut(H)\times Aut(K).

Proof:

Now suppose G=HK, where H and K are characteristic subgroups of G with H\cap K=\{e\}. Define \Psi:\text{Aut}(G)\to\text{Aut}(H)\times\text{Aut}(K) by \displaystyle \Psi(\sigma)=(\sigma|_H, \sigma|_K).

\sigma|_H:H\to H is a homomorphism, and bijective since \sigma|_H(H)=H. Thus \sigma|_H\in\text{Aut}(H) and similarly, \sigma|_K\in\text{Aut}(K) so that \Psi is well-defined.

Note that \displaystyle \Psi(\sigma_1\sigma_2)=((\sigma_1\sigma_2)|_H, (\sigma_1\sigma_2)|_K)=(\sigma_1|_H,\sigma_1|_K)(\sigma_2|_H,\sigma_2|_K)=\Psi(\sigma_1)\Psi(\sigma_2) so \Psi is a homomorphism.

Suppose \sigma\in\ker\Psi. Then \Psi(\sigma)=(\sigma|_H,\sigma|_K)=(\text{id}_H,\text{id}_K). Then for hk\in G, \sigma(hk)=\sigma(h)\sigma(k)=hk so that \sigma=\text{id}_G. Thus \Psi is injective.

For any (\phi, \psi)\in\text{Aut}(H)\times\text{Aut}(K), define \sigma(hk)=\phi(h)\psi(k). Then
\begin{aligned} \sigma(h_1k_1h_2k_2)&=\sigma(h_1h_2k_1k_2)\\ &\text{(}H, K\ \text{normal and}\ H\cap K=\{e\}\ \text{implies elements of}\ H, K\ \text{commute)}\\ &=\phi(h_1h_2)\psi(k_1k_2)\\ &=\phi(h_1)\phi(h_2)\psi(k_1)\psi(k_2)\\ &=\phi(h_1)\psi(k_1)\phi(h_2)\psi(k_2)\\ &=\sigma(h_1k_1)\sigma(h_2k_2). \end{aligned}
So \sigma is a homomorphism.

If hk\in\ker\sigma, then \phi(h)\psi(k)=e, so that \phi(h)=(\psi(k))^{-1}. Then since H\cap K=\{e\}, so \phi(h)=\psi(k)=e, so that h=k=e. Thus \ker\sigma=\{e\} and \sigma is injective.

Any h\in H can be written as \phi(h') since \phi is bijective. Similarly, any k\in K can be written as \psi(k'). Then \sigma(h'k')=\phi(h')\psi(k')=hk so \sigma is surjective.

Thus \sigma\in\text{Aut}(G). Note that \sigma|_H=\phi since \sigma|_H(h)=\sigma(h\cdot 1)=\phi(h)\psi(1)=\phi(h). Similarly, \sigma|_K=\psi. So \Psi(\sigma)=(\phi,\psi) and \Psi is surjective.

Hence \Psi is an isomorphism.

How to Remember the 8 Vector Space Axioms

Vector Space has a total of 8 Axioms, most of which are common-sense, but can still pose a challenge for memorizing by heart.

I created a mnemonic “MAD” which helps to remember them.

M for Multiplicative Axioms:

  1. 1x=x (Scalar Multiplication identity)
  2. (ab)x=a(bx) (Associativity of Scalar Multiplication)

A for Additive Axioms: (Note that these are precisely the axioms for an abelian group)

  1. x+y=y+x (Commutativity)
  2. (x+y)+z=x+(y+z) (Associativity for Vector Addition)
  3. x+(-x)=0 (Existence of Additive Inverse)
  4. x+0=0+x=0 (Additive Identity)

D for Distributive Axioms:

  1. a(x+y)=ax+ay (Distributivity of vector sums)
  2. (a+b)x=ax+bx (Distributivity of scalar sums)

How to Remember the 10 Field Axioms

There are a total of 10 Axioms for Field, it can be quite a challenge to remember all 10 of them offhand.

I created a mnemonic “ACIDI” to remember the 10 axioms. Unfortunately it is not a real word, but is close to the word “acidic”. A picture to remember is “acidic field”, a grass field polluted by acid rain?! 😛

A: Associativity
C: Commutativity
I: Identity
D: Distributivity
I: Inverse

Each of the properties has two parts – Addition and Multiplication. This table from Wolfram summarizes it perfectly:

name addition multiplication
associativity (a+b)+c=a+(b+c) (ab)c=a(bc)
commutativity a+b=b+a ab=ba
distributivity a(b+c)=ab+ac (a+b)c=ac+bc
identity a+0=a=0+a a·1=a=1·a
inverses a+(-a)=0=(-a)+a aa^(-1)=1=a^(-1)a if a!=0

Finite group generated by two elements of order 2 is isomorphic to Dihedral Group

Suppose G=\langle s,t\rangle where both s and t has order 2. Prove that G is isomorphic to D_{2m} for some integer m.

Note that G=\langle st, t\rangle since (st)t=s. Since G is finite, st has a finite order, say m, so that (st)^m=1_G. We also have [(st)t]^2=s^2=1.

We claim that there are no other relations, other than (st)^m=t^2=[(st)t]^2=1.

Suppose to the contrary sts=1. Then sstss=ss, i.e. t=1, a contradiction. Similarly if ststs=1, tsststsst=tsst implies s=1, a contradiction. Inductively, (st)^ks\neq 1 and (ts)^kt\neq 1 for any k\geq 1.

Thus \displaystyle G\cong D_{2m}=\langle a,b|a^m=b^2=(ab)^2=1\rangle.

Three Properties of Galois Correspondence

The Fundamental Theorem of Galois Theory states that:

Given a field extension E/F that is finite and Galois, there is a one-to-one correspondence between its intermediate fields and subgroups of its Galois group.
1) H\leftrightarrow E^H where H\leq\text{Gal}(E/F) and E^H is the corresponding fixed field (the set of those elements in E which are fixed by every automorphism in H).
2) K\leftrightarrow\text{Aut}(E/K) where K is an intermediate field of E/F and \text{Aut}(E/K) is the set of those automorphisms in \text{Gal}(E/F) which fix every element of K.

This correspondence is a one-to-one correspondence if and only if E/F is a Galois extension.

Three Properties of the Galois Correspondence

  1. It is inclusing-reversing. The inclusion of subgroups H_1\subseteq H_2 holds iff the inclusion of fields E^{H_2}\subseteq E^{H_1} holds.
  2. If H is a subgroup of \text{Gal}(E/F), then |H|=[E:E^H] and |\text{Gal}(E/F)/H|=[E^H:F].
  3. The field E^H is a normal extension of F (or equivalently, Galois extension, since any subextension of a separable extension is separable) iff H is a normal subgroup of \text{Gal}(E/F).

Characterization of Galois Extensions

Characterization of Galois Extensions

For a finite extension E/F, each of the following statements is equivalent to the statement that E/F is Galois:

1) E/F is a normal extension and a separable extension.
2) Every irreducible polynomial in F[x] with at least one root in E splits over E and is separable.
3) E is a splitting field of a separable polynomial with coefficients in F.
4) |\text{Aut}(E/F)|=[E:F], that is, the number of automorphisms equals the degree of the extension.
5) F is the fixed field of \text{Aut}(E/F).

Fundamental Theorem of Galois Theory

Given a field extension E/F that is finite and Galois, there is a one-to-one correspondence between its intermediate fields and subgroups of its Galois group.
H\leftrightarrow E^H

where H\leq\text{Gal}(E/F) and E^H is the corresponding fixed field (the set of those elements in E which are fixed by every automorphism in H).
K\leftrightarrow\text{Aut}(E/K)

where K is an intermediate field of E/F and \text{Aut}(E/K) is the set of those automorphisms in \text{Gal}(E/F) which fix every element of K.

This correspondence is a one-to-one correspondence if and only if E/F is a Galois extension.

Examples
1) E\leftrightarrow\{\text{id}_E\}, the trivial subgroup of \text{Gal}(E/F).
2) F\leftrightarrow\text{Gal}(E/F).

Class Equation of a Group

The class equation of a group is something that looks difficult at first sight, but is actually very straightforward once you understand it. An amazing equation…

Class Equation of a Group (Proof)

Suppose G is a finite group, Z(G) is the center of G, and c_1, c_2, \dots, c_r are all the conjugacy classes in G comprising the elements outside the center. Let g_i be an element in c_i for each 1\leq i\leq r. Then we have: \displaystyle |G|=|Z(G)|+\sum_{i=1}^r[G:C_G(g_i)].

Proof:

Let G act on itself by conjugation. The orbits of G partition G. Note that each conjugacy class c_i is actually \text{Orb}(g_i).

Let x\in Z(G). Then gxg^{-1}=xgg^{-1}=x for all g\in G. Hence \text{Orb}(x) consists of a single element x itself.

Let g_i\in c_i. Then
\begin{aligned} \text{Stab}(g_i)&=\{h\in G\mid hg_ih^{-1}=g_i\}\\ &=\{h\in G\mid hg_i=g_ih\}\\ &=C_G(g_i). \end{aligned}
By Orbit-Stabilizer Theorem, \displaystyle |\text{Orb}(g_i)|=[G:\text{Stab}(g_i)]=[G:C_G(g_i)].

Therefore, \displaystyle |G|=|Z(G)|+\sum_{i=1}^r[G:C_G(g_i)].

Orbit-Stabilizer Theorem (with proof)

Orbit-Stabilizer Theorem

Let G be a group which acts on a finite set X. Then \displaystyle |\text{Orb}(x)|=[G:\text{Stab}(x)]=\frac{|G|}{|\text{Stab}(x)|}.

Proof

Define \phi:G/\text{Stab}(x)\to\text{Orb}(x) by \displaystyle \phi(g\text{Stab}(x))=g\cdot x.

Well-defined:

Note that \text{Stab}(x) is a subgroup of G. If g\text{Stab}(x)=h\text{Stab}(x), then g^{-1}h\in\text{Stab}(x). Thus g^{-1}hx=x, which implies hx=gx, thus \phi is well-defined.

Surjective:

\phi is clearly surjective.

Injective:

If \phi(g\text{Stab}(x))=\phi(h\text{Stab}(x)), then gx=hx. Thus g^{-1}hx=x, so g^{-1}h\in\text{Stab}(x). Thus g\text{Stab}(x)=h\text{Stab}(x).

By Lagrange’s Theorem, \displaystyle \frac{|G|}{|\text{Stab}(x)|}=|G/\text{Stab}(x)|=|\text{Orb}(x)|.

Field Medallist Prof. Gowers has also written a nice post on the Orbit -Stabilizer Theorem and various proofs.

Necessary and Sufficient Conditions for Semidirect Product to be Abelian (Proof)

This theorem is pretty basic, but it is useful to construct non-abelian groups. Basically, once you have either group to be non-abelian, or the homomorphism to be trivial, the end result is non-abelian!

Theorem: The semidirect product N\rtimes_\varphi H is abelian iff N, H are both abelian and \varphi: H\to\text{Aut}(N) is trivial.

Proof:
(\implies)

Assume N\rtimes_\varphi H is abelian. Then for any n_1, n_2\in N, h_1, h_2\in H, we have
\begin{aligned} (n_1, h_1)\cdot(n_2,h_2)&=(n_2,h_2)\cdot(n_1, h_1)\\ (n_1\varphi_{h_1}(n_2), h_1h_2)&=(n_2\varphi_{h_2}(n_1), h_2h_1). \end{aligned}
This implies h_1h_2=h_2h_1, thus H is abelian.

Consider the case n_1=n_1=n. Then for any n\in N, n\varphi_{h_1}(n)=n\varphi_{h_2}(n). Multiplying by n^{-1} on the left gives \varphi_{h_1}(n)=\varphi_{h_2}(n) for any h_1, h_2\in H. Thus \varphi_h(n)=\varphi_e(n)=n for all h\in H so \varphi is trivial.

Consider the case where h_1=h_2=e. Then we have n_1n_2=n_2n_1, so N has to be abelian.

(\impliedby)

This direction is clear.

On Semidirect Products

Outer Semidirect Product

Given any two groups N and H and a group homomorphism \phi:H\to\text{Aut}(N), we can construct a new group N\rtimes_\phi H, called the (outer) semidirect product of N and H with respect to \phi, defined as follows.
(i) The underlying set is the Cartesian product N\times H.
(ii) The operation, \bullet, is determined by the homomorphism \phi:

\bullet: (N\rtimes_\phi H)\times (N\rtimes_\phi H)\to N\rtimes_\phi H

(n_1,h_1)\cdot(n_2,h_2)=(n_1\phi_{h_1}(n_2),h_1h_2)

for n_1,n_2\in N and h_1,h_2\in H.

This defines a group in which the identity element is (e_N, e_H) and the inverse of the element (n,h) is (\phi_{h^{-1}}(n^{-1}), h^{-1}).
Pairs (n, e_H) form a normal subgroup isomorphic to N, while pairs (e_N, h) form a subgroup isomorphic to H.

Inner Semidirect Product (Definition)

Given a group G with identity element e, a subgroup H, and a normal subgroup N\lhd G; then the following statements are equivalent:

(i) G is the product of subgroups, G=NH, where the subgroups have trivial intersection, N\cap H=\{e\}.
(ii) For every g\in G, there are unique n\in N and h\in H, such that g=nh.

If these statements hold, we define G to be the semidirect product of N and H, written G=N\rtimes H.

Inner Semidirect Product Implies Outer Semidirect Product

Suppose we have a group G with N\lhd G, H\leq G and every element g\in G can be written uniquely as g=nh where n\in N, h\in H.

Define \phi: H\to\text{Aut}(N) as the homomorphism given by \phi(h)=\phi_h, where \phi_h(n)=hnh^{-1} for all n\in N, h\in H.

Then G is isomorphic to the semidirect product N\rtimes_{\phi}H, and applying the isomorphism to the product, nh, gives the tuple, (n,h). In G, we have
\displaystyle (n_1h_1)(n_2h_2)=n_1h_1n_2(h_1^{-1}h_1)h_2=(n_1\phi_{h_1}(n_2))(h_1h_2)=(n_1,h_1)\cdot(n_2,h_2)
which shows that the above map is indeed an isomorphism.

Sylow Theorems

Sylow Theorems

Let G be a finite group.

Theorem 1

For every prime factor p with multiplicity n of the order of G, there exists a Sylow p-subgroup of G, of order p^n.

Theorem 2

All Sylow p-subgroups of G are conjugate to each other, i.e.\ if H and K are Sylow p-subgroups of G, then there exists an element g\in G with g^{-1}Hg=K.

Theorem 3

Let p be a prime such that |G|=p^nm, where p\nmid m. Let n_p be the number of Sylow p-subgroups of G. Then:
1) n_p\mid m, which is the index of the Sylow p-subgroup in G.
2) n_p\equiv 1\pmod p.

Theorem 3b (Proof)

We have n_p=[G:N_G(P)], where P is any Sylow p-subgroup of G and N_G denotes the normalizer.

Proof

Let P be a Sylow p-subgroup of G and let G act on \text{Syl}_p(G) by conjugation. We have |\text{Orb}(P)|=n_p, \text{Stab}(P)=\{g\in G:gPg^{-1}=P\}=N_G(P).

By the Orbit-Stabilizer Theorem, |\text{Orb}(P)|=[G:\text{Stab}(P)], thus n_p=[G:N_G(P)].

Orbit-Stabilizer Theorem

Let G be a group which acts on a finite set X. Then \displaystyle |\text{Orb}(x)|=[G:\text{Stab}(x)]=\frac{|G|}{|\text{Stab}(x)|}.

FTFGAG: Fundamental Theorem of Finitely Generated Abelian Groups

Fundamental Theorem of Finitely Generated Abelian Groups

Primary decomposition

Every finitely generated abelian group G is isomorphic to a group of the form \displaystyle \mathbb{Z}^n\oplus\mathbb{Z}_{q_1}\oplus\dots\oplus\mathbb{Z}_{q_t} where n\geq 0 and q_1,\dots,q_t are powers of (not necessarily distinct) prime numbers. The values of n, q_1, \dots, q_t are (up to rearrangement) uniquely determined by G.

Invariant factor decomposition

We can also write G as a direct sum of the form \displaystyle \mathbb{Z}^n\oplus\mathbb{Z}_{k_1}\oplus\dots\oplus\mathbb{Z}_{k_u}, where k_1\mid k_2\mid k_3\mid\dots\mid k_u. Again the rank n and the invariant factors k_1,\dots,k_u are uniquely determined by G.

Finite extension is Algebraic extension (Proof) + “Converse”

These two are useful lemmas in Galois/Field Theory.

Finite extension is Algebraic extension (Proof)
Let L/K be a finite field extension. Then L/K is an algebraic extension.
Proof:
Let L/K be a finite extension, where [L:K]=n. Let \alpha\in L. Consider \{1,\alpha,\alpha^2,\dots,\alpha^n\} which has to be linearly dependent over K since there are n+1 elements. Thus, there exists c_i\in K (not all zero) such that \sum_{i=0}^n c_i\alpha^i=0, so \alpha is algebraic over K.

Finitely Generated Algebraic Extension is Finite (Proof)
Let L/K be a finitely generated algebraic extension. Then L/K is a finite extension.

Proof:
Since L/K is finitely generated, L=K(\alpha_1,\dots,\alpha_n) for some \alpha_1,\dots,\alpha_n\in K. Since L/K is algebraic, each \alpha_i is algebraic over K. Denote L_i:=K(\alpha_1,\dots,\alpha_i) for 1\leq i\leq n. Then L_i=L_{i-1}(\alpha_i) for each i. Since \alpha_i is algebraic over K, it is also algebraic over L_{i-1}, so there exists a polynomial g_i with coefficients in L_{i-1} such that g_i(\alpha_i)=0. Thus [L_i:L_{i-1}]\leq\deg g_i<\infty. Similarly [L_1:K]<\infty. By Tower Law, [L:K]=[L_n:L_{n-1}][L_{n-1}:L_{n-2}]\dots[L_1:K]<\infty.

Endomorphism ring of Q is a division algebra

We show that Q is not semisimple nor simple, but \text{End}_\mathbb{Z}(\mathbb{Q}) is a division algebra.

Consider A=\mathbb{Z} (as a \mathbb{Z}-algebra). Consider M=\mathbb{Q} as a right \mathbb{Z}-module.
Lemma:
\mathbb{Q} is not semisimple nor simple.

Suppose to the contrary \mathbb{Q}=\bigoplus_{i\in I}N_i, where N_i are simple \mathbb{Z}-modules (i.e. N_i\cong\mathbb{Z}/p_i\mathbb{Z}). Then there exists nonzero x\in\mathbb{Q} such that x has finite order (product of primes). This is impossible in \mathbb{Q}.
Lemma:
\text{End}_\mathbb{Z}(\mathbb{Q})\cong\mathbb{Q} as \mathbb{Z}-algebras.

Define \Psi:\mathbb{Q}\to\text{End}_\mathbb{Z}(\mathbb{Q}) where q\in\mathbb{Q} is mapped to \lambda_q\in\text{End}_\mathbb{Z}(\mathbb{Q}), where \lambda_q(x)=qx. Let k\in\mathbb{Z}, q,q_1,q_2\in\mathbb{Q}.

We can check that \Psi is a \mathbb{Z}-algebra homomorphism.

Let q\in\ker\Psi. Then \Psi(q)=\lambda_q=0, \lambda_q(x)=qx=0 for all x\in\mathbb{Q}. This implies q=q\cdot 1=0. Hence \Psi is injective.

Let \phi\in\text{End}_\mathbb{Z}(\mathbb{Q}). Let x=\frac{a}{b}\in\mathbb{Q}, where a,b\in\mathbb{Z}. \phi(x)=a\phi(\frac 1b)=\frac ab\cdot b\phi(\frac 1b)=\frac ab\cdot\phi(1)=\phi(1)\cdot x=\lambda_{\phi(1)}(x). Hence \Psi is surjective.

Thus \text{End}_\mathbb{Z}(\mathbb{Q})\cong\mathbb{Q} is a division algebra, but \mathbb{Q} is not simple.

Artin-Whaples Theorem

There seems to be another version of Artin-Whaples Theorem, called the Artin-Whaples Approximation theorem.

The theorem stated here is Artin-Whaples Theorem for central simple algebras.

Artin-Whaples Theorem: Let A be a central simple algebra over a field F. Let a_1,\dots,a_n\in A be linearly independent over F and let b_1,\dots,b_n be any elements in A. Then there exists a_i',a_i''\in A for i=1,\dots,m such that the F-linear map f:A\to A defined by f(x)=\sum_{r=1}^m a_r'xa_r'' satisfies f(a_j)=b_j for all j=1,\dots,n.

Very nice and useful theorem.

Simple Algebra does not imply Semisimple Algebra

The terminology “semisimple” algebra suggests a generalization of simple algebras, but in fact not all simple algebras are semisimple! (Exercises 1 & 5 in Richard Pierce’s book contain examples)

A simple module is a semisimple module is true though.

Proposition: For a simple algebra A, the following conditions are equivalent:

(i) A is semisimple;

(ii) A is right Artinian;

(iii) A has a minimal right ideal.

Thus to find a algebra that is simple but not semisimple, one can look for an example that is not right Artinian.

Wedderburn’s Structure Theorem

In abstract algebra, the Artin–Wedderburn theorem is a classification theorem for semisimple rings and semisimple algebras. The theorem states that an (Artinian) [1] semisimple ring R is isomorphic to a product of finitely many ni-by-ni matrix rings over division rings Di, for some integers ni, both of which are uniquely determined up to permutation of the index i. In particular, any simple left or right Artinian ring is isomorphic to an n-by-nmatrix ring over a division ring D, where both n and D are uniquely determined. (Wikipedia)

This is quite a powerful theorem, as it allows semisimple rings/algebras to be “represented” by a finite direct sum of matrix rings over division rings. This is in the spirit of Representation theory, which tries to convert algebraic objects into objects in linear algebra, which is relatively well understood.

Wedderburn’s Structure Theorem

Let A be a semisimple R-algebra.

(i) A\cong M_{n_1}(D_1)\oplus\dots\oplus M_{n_r}(D_r) for some natural numbers n_1,\dots,n_r and R-division algebras D_1,\dots,D_r.

(ii) The pair (n_i,D_i) is unique up to isomorphism and order of arrangement.

(iii) Conversely, suppose A=M_{n_1}(D_1)\oplus\dots\oplus M_{n_r}(D_r), then A is a right (and left) semisimple R-algebra.

Some Notes

The definition of a semisimple R-algebra is: An R-algebra A is semisimple if A is semisimple as a right A-module.

Example: R=\mathbb{R} and A=M_2(\mathbb{R})\oplus M_2(\mathbb{R}). Then A is semisimple by Wedderburn’s theorem.

Semisimple Modules Equivalent Conditions

Proposition:  For a right A-module M, the following are equivalent:

(i) M is semisimple.

(ii) M=\sum\{N\in S(M): N\ \text{is simple}\}.

(iii) S(M) is a complemented lattice, that is, every submodule of M has a complement in S(M).

We are following Pierce’s book’s Associative Algebras (Graduate Texts in Mathematics) notation, where S(M) is the set of all submodules of M.

This proposition is can be used to prove two useful Corollaries:

Corollary 1) If M is semisimple and P is a submodule of M, then both P and M/P are semisimple. In words, this means that submodules and quotients of a semisimple module is again semisimple.

Proof: Since M is semisimple, by (iii) we have M\cong P\oplus P', where P'\cong M/P. Let N be a submodule of P\leq M. Then N has a complement N' in S(M): M=N\oplus N'.

P=P\cap(N\oplus N')=N\oplus (N'\cap P) with N'\cap P\in S(P). Thus, we have that P is semisimple by condition (iii). Similarly, M/P\cong P' is semisimple.

Corollary 2) A direct sum of semisimple modules is semisimple.

Proof: This is quite clear from the definition of semisimple modules being direct sum of simple modules. A direct sum of (direct sum of simple modules) is again a direct sum of simple modules.