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Gauss's Lemma for polynomials claims that a non-constant polynomial in $\mathbb{Z}[X]$ is irreducible in $\mathbb{Z}[X]$ if and only if it is both irreducible in $\mathbb{Q}[X]$ and primitive in $\mathbb{Z}[X]$.

I wonder if this holds for multivariable case; is it true that a non-constant polynomial in $\mathbb{Z}[X_1,\dots,X_n]$ is irreducible in $\mathbb{Z}[X_1,\dots,X_n]$ if and only if it is both irreducible in $\mathbb{Q}[X_1,\dots,X_n]$ and primitive in $\mathbb{Z}[X_1,\dots,X_n]$?

Thank you for your help.

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Some generalizations of Gauss' Lemma are discussed at's_lemma_(polynomial) and you might find that what you want is there. – Gerry Myerson Oct 14 '12 at 8:06

Let $n \ge 1$ be an integer. We define a total order on $\mathbb{Z}^n$ as follows. Let $(r_1,\dots, r_n)$, $(s_1,\dots, s_n) \in \mathbb{Z}^n$. Let $k = \min \{i; r_i \neq s_i\}$. Then $(r_1,\dots, r_n) > (s_1,\dots, s_n)$ if and only if $r_k > s_k$.

Lemma 1 Let $r, s, t \in \mathbb{Z}^n$. Suppose $r > s$. Then $r + t > s + t$.

Proof: Clear.

Lemma 2 Let $r, s, r', s' \in \mathbb{Z}^n$. Suppose $r + s = r' + s'$ and $r > r'$. Then $s' > s$.

Proof: $r - r' = s' - s$. By Lemma 1, $r - r' > 0$. Hence $s' - s > 0$. Hence $s' > s$ by Lemma 1. QED

Let $A$ be a UFD. Let $p$ be a prime element of $A$. Let $x \in A$. If $x$ is divisible by $p^a$ but not by $p^{a+1}$, we denote this fact by $p^a||x$.

Let $f \in A[X_1,\dots, X_n]$. We denote by $C(f)$ the gcd of all the coefficients of $f$. If $(C(f)) = (1)$, $f$ is called primitive.

Let $\mathbb{N}$ be the set of integers $\ge 0$. We denote $(r_1,\dots, r_n) \in \mathbb{N}^n$ by $r$. We denote a monomial $X_1^{r_1}\cdots X_n^{r_n}$ by $X^r$.

Lemma 3 Let $A$ be a UFD. Let $f, g \in A[X_1,\dots, X_n]$. Then $(C(fg)) = (C(f)C(g))$.

Proof: Let $f = \sum_r \lambda_r X^r$. Let $g = \sum_s \mu_s X^s$. Let $fg = \sum_m \gamma_m X^m$. Then $\gamma_m = \sum_{r+s = m} \lambda_r\mu_s$. Let $p$ be a prime element of $A$. Suppose $p^a||C(f)$ and $p^b||C(g)$. It suffices to prove that $p^{a+b}||C(fg)$. It is clear that $p^{a+b}|C(fg)$. Let $h = max \{r \in \mathbb{N}^n\colon \lambda_r$ is not divisible by $p^{a+1}\}$. Let $k = max \{s \in \mathbb{N}^n\colon \mu_s$ is not divisible by $p^{b+1}\}$. Let $r, s \in \mathbb{N}^n$. Suppose $r + s = h + k$. If $r \neq h$, then $\lambda_r\mu_s$ is divisible by $p^{a+b+1}$ by Lemma 2. Since $\gamma_{h+k} = \sum_{r+s = h+k} \lambda_r\mu_s$, $\gamma_{h+k}$ is not divisible by $p^{a+b+1}$. QED

Proposition Let $A$ be a UFD, $K$ its field of fractions. Let $f \in A[X_1,\dots, X_n]$ be non-constant. Then $f$ is irreducible if and only if $f$ is primitive and $f$ is irreducible in $K[X_1,\dots, X_n]$.

Proof: Suppose $f$ is irreducible. Clearly $f$ is primitive. Suppose $f = g'h'$, where $g'$ and $h'$ are non-constant polynomial in $K[X_1,\dots, X_n]$. It is easy to see that there exist primitive polynomials $g, h \in A[X_1,\dots, X_n]$ and $a, b \in A -\{0\}$ such that $af = bgh$. By Lemma 3, $gh$ is primitive. Hence $b = a\epsilon$, where $\epsilon$ is an invertible element of $A$. Hence $f = \epsilon gh$. This is a contradiction.

The converse is clear. QED

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Maybe one should say that you use the same idea of proof from the case of a single indeterminate and the lexicographical order for monomials instead of the ordering given by degree. – user26857 Oct 17 '12 at 8:54
@navigetor23 That's about it. – Makoto Kato Oct 17 '12 at 10:01

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