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.

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    $\begingroup$ "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? $\endgroup$ Dec 15, 2022 at 21:27
  • $\begingroup$ P.S.: "Gauss's". See for example the Chicago Manual of Style. $\endgroup$ Dec 15, 2022 at 21:39
  • $\begingroup$ @ArturoMagidin yes it should, I've edited it now. Thanks :) $\endgroup$ Dec 15, 2022 at 21:58

1 Answer 1


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.

  • $\begingroup$ "(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? $\endgroup$ Dec 15, 2022 at 22:59
  • $\begingroup$ If I'm not mistaken, this proof strategy works in any GCD domain, which is nice $\endgroup$ Dec 15, 2022 at 23:02
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    $\begingroup$ @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). $\endgroup$ Dec 16, 2022 at 1:45
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    $\begingroup$ @ducksforever See the extensive survey by D.D. Anderson cited here for generalizations of Gauss's Lemma and related results. $\endgroup$ Dec 19, 2022 at 10:15

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