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 $D[x]$.