# Two forms of Gauss' lemma

There are two statements which often get called Gauss' lemma

Gauss' lemma version 1: Let $$f,g\in \mathbb{Z}[x]$$. If $$f$$,$$g$$ are both primitive, then $$fg$$ is primitive.

Gauss' lemma version 2: Let $$f \in \mathbb{Z}[x]$$. Then $$f$$ is irreducible in $$\mathbb{Z}[x]$$ if and only if is is primitive and is irreducible in $$\mathbb{Q}[x]$$.

It is pretty straightforward to see that version 2 follows from version 1. However, I am not familiar with any argument by which you can prove version 1, assuming you know version 2. Is there any such argument, or is version 2 better thought of as a consequence of version 1?

Definitions: We say that a polynomial in $$\mathbb{Z}[x]$$ is primitive if the greatest common divisor of its coefficients is 1.

I am aware that there of course is a trivial argument: assume version 2, then give the usual proof of version 1. I am looking for an argument which isn't of this sort.

• "I am not familiar with any argument by which you can prove version 1, assuming you know version 1". Should the second 1 be a 2? Dec 15, 2022 at 21:27
• P.S.: "Gauss's". See for example the Chicago Manual of Style. Dec 15, 2022 at 21:39
• @ArturoMagidin yes it should, I've edited it now. Thanks :) Dec 15, 2022 at 21:58

What Gauss actually proved was what you give as Version 2, in contrapositive form. In Chapter II, article 42 of the Disquisitiones Arithmeticae, he writes:

If the coefficients $$A$$, $$B$$, $$C,\ldots, N$$; $$a$$, $$b$$, $$c,\ldots,n$$ of two functions of the form \begin{align*} x^m+Ax^{m-1}+Bx^{m-2}+Cx^{m-3}\ldots + N \tag{P}\\ x^n + ax^{n-1}+bx^{n-2}+cx^{n-3}\ldots + n \tag{Q} \end{align*} are all rational and not all integers, and if the product of (P) and (Q) $$= x^{m+n}+\mathfrak{A}x^{m+n-1} + \mathfrak{B}x^{m+n-2} + \text{etc.} + \mathfrak{Z}$$ then not all the coefficients $$\mathfrak{A},\mathfrak{B},\ldots\mathfrak{Z}$$ can be integers.

See discussions here and here. Version 2 is certainly a consequence of Version 1, as you note.

To obtain Version 1 from Version 2, first we factor $$f(x)$$ and $$g(x)$$ into irreducibles in $$\mathbb{Z}[x]$$ (the factorization exists, though we do not yet know it is unique): \begin{align*} f(x) &= f_1(x)\cdots f_r(x)\\ g(x) &= g_1(x)\cdots g_s(x). \end{align*} Since the product of the contents certainly divides the content of the product, we know that each $$f_i$$ and $$g_j$$ is primitive and therefore irreducible in $$\mathbb{Q}[x]$$ by Version 2.

Let $$c$$ be the content of $$f(x)g(x)$$, and write $$f(x)g(x) = cF(x)$$ with $$F(x)\in\mathbb{Z}[x]$$ primitive. Since $$F(x)$$ is an associate of $$f(x)g(x)$$ in $$\mathbb{Q}[x]$$, it is reducible in $$\mathbb{Q}[x]$$, hence in $$\mathbb{Z}[x]$$; we can likewise factor $$F(x)$$ into irreducibles in $$\mathbb{Z}[x]$$, all of them primitive, $$F(x) = F_1(x)\cdots F_t(x).$$ So we have in $$\mathbb{Q}[x]$$ that $$cF_1(x)\cdots F_t(x) = f_1(x)\cdots f_r(x)g_1(x)\cdots g_s(x).$$ Because each nonconstant factor is irreducible in $$\mathbb{Z}[x]$$ and primitive, they are irreducible in $$\mathbb{Q}[x]$$ by Version 2. By unique factorization in $$\mathbb{Q}[x]$$, it follows that $$t=r+s$$, and we can reorder and relabel the irreducible factors of $$F(x)$$ as $$F(x) = F_1(x)\cdots F_r(x)G_1(x)\cdots G_s(x)$$ where each $$F_i$$ and $$G_j$$ has integer coefficients and is primitive; and moreover each $$F_i(x)$$ is a $$\mathbb{Q}$$-associate of the corresponding $$f_i(x)$$, and each $$G_j(x)$$ is a $$\mathbb{Q}$$-associate of $$g_j(x)$$.

But now assume that $$a(x)$$ and $$b(x)$$ are nonzero, primitive with integer coefficients, and they are $$\mathbb{Q}$$-associates. Then there exists integers $$m$$ and $$n$$, $$\gcd(m,n)=1$$, with $$\frac{m}{n}a(x) = b(x)$$. Then $$ma(x) = nb(x)$$. Since the content of $$ma(x)$$ is $$|m|$$, and the content of $$nb(x)$$ is $$|n|$$, then $$|m|=|n|=1$$, so $$\frac{m}{n}=\pm 1$$. Thus, $$a(x)$$ and $$b(x)$$ are either equal, or negatives of each other.

That means that $$f_i(x)=\pm F_i(x)$$ and $$g_j(x)=\pm G_j(x)$$ for each $$i$$ and $$j$$. Thus, we have \begin{align*} f(x)g(x) &= cF(x) \\ &= cF_1(x)\cdots F_r(x)G_1(x)\cdots G_s(x)\\ &= \pm c f_1(x)\cdots f_r(x)g_1(x)\cdots g_s(x)\\ &= \pm c f(x)g(x). \end{align*} Therefore, $$|c|=1$$, proving that the product of primitive polynomials is primitive.

• "(the factorisation exists, although we do not yet know it is unique" - is this just because for any domain $R$, any element $f \in R[x]$ can be factored into irreducibles by degree considerations? Dec 15, 2022 at 22:59
• If I'm not mistaken, this proof strategy works in any GCD domain, which is nice Dec 15, 2022 at 23:02
• @ducksforever In this case, yes, degree considerations, plus factorization of integers, tells you that you can do it. You wouldn't be able to guarantee such a factorization into irreducibles if you don't have it in the domain itself. For instance, over the algebraic integers you cannot factor constants into irreducibles. I would be careful about gcd domains, though... The algebraic integeres are a gcd domain (in fact a Bezout domain). Dec 16, 2022 at 1:45
• @ducksforever See the extensive survey by D.D. Anderson cited here for generalizations of Gauss's Lemma and related results. Dec 19, 2022 at 10:15