Gauss' lemma: $k(x_1, \ldots, x_n)[t]$ and $k[x_1,\ldots,x_n][t]$.

Question

Let $f(t)$ be an irreducible polynomial in $k(x_1, \ldots, x_n)[t]$. Show that $f(t)$ is irreducible in $k[x_1,\ldots,x_n][t]$.

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I guess it's clear that I am required to apply Gauss' Lemma. The statement in my course notes is:

Let $R$ be a UFD and $F$ its field of fractions. A non-constant polynomial in $R[X]$ is irreducible in $R[X]$ if and only if it is both irreducible in $F[X]$ and primitive in $R[X]$. And a polynomial $P$ in $R[X]$ is primitive if the only elements of $R$ that divide all coefficients of $P$ at once are the invertible elements of $R$.

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I first cleared the denominators of the coefficients $a_i \in k(x_1,\ldots,x_n)$ of $f(t)$. But this modified polynomial is not monic anymore, so I ran into trouble showing that $f$ is primitive in $k[x_1,\ldots,x_n][t]$ - or at least, things get very messy.

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EDIT: Additional context added. I am trying to understand Proposition 4.9 in Hartshorne's Algebraic Geometry. The argument given goes as follows:

• Let $K$ be a finitely generated extension field of $k$. Then $K$ is seperably generated over $k$ since we are working over an algebraically closed field. Hence we can find a transcendence base $x_1,\ldots,x_n \in K$ such that $K$ is a finite seperable extension of $k(x_1,\ldots,x_n)$. By the primitive element therem we can find one further element $y \in K$ such that $K=k(x_1,\ldots,x_n,y)$. Now $y$ is algebraic over $k(x_1,\ldots,x_n)$. So it satisfies a polynomial equation with coefficients which are rational functions in $x_1,\ldots,x_n$. Clearing denominators we get an irreducible polynomial, satisfying $f(x_1,\ldots,x_n,y)=0$

• My original question aimed to clarify this last step. I assumed he used Gauss' Lemma.

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EDIT 2: Reformulation of original question (related to comments on first post).

Let $f(t)$ be an irreducible polynomial in $k(x_1, \ldots, x_n)[t]$. Let $f'(t)$ be the polynomial where we have cleared up the denominators of $f(t)$. Show that $f'(t)$ is irreducible in $k[x_1,\ldots,x_n][t]$.

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Answer 1

From the comments, it seems like you want to ask the following question. Suppose $f(t) \in k(x_1,...,x_n)[t]$ is irreducible (where $k$ is a field). We can assume that the coefficients are reduced so that for each coefficient $\frac{c_i}{d_i}$, the numerator and denominator are relatively prime. Then $k[x_1, ..., x_n]$ is a UFD so we can take the least common multiple $m$ of the denominators of the coefficients of $f(t)$ and gcd $n$ of all the numerators of the coefficients. Let $f'(t)=\frac{m}{n}f(t) \in k[x_1,...,x_n][t]$. We are in an integral domain and $\frac{m}{n} \neq 0$ so the roots of $f'$ are the same as the roots of $f$. If $f$ is irreducible in $k(x_1, ..., x_n)[t]$, is $f'(t)$ irreducible in $k[x_1,...,x_n][t]$?

Suppose $p$ is a prime factor dividing each coefficient $\frac{m}{d_i} \cdot \frac{c_i}{n}$ of $f'$. Then for each $i$, $p|\frac{m}{d_i}$ or $p|\frac{c_i}{n}$. If in all cases the latter is true, then $p| c_i$ for each $i$ so this contradicts the definition of $n$. Thus, $p|\frac{m}{d_i}$ for some $i$, and in particular, $p|m$. Thus, if $d_i$ contains the largest power of $p$ of all the denominators, then $p$ does not divide $\frac{m}{d_i}$ so must divide $\frac{c_i}{n}$. But then $p|c_i$ which contradicts $p|d_i$ (and we assumed $\frac{c_i}{d_i}$ is reduced). Thus, $f'$ is primitive and so if we can show it is irreducible in $k(x_1, ..., x_n)[t]$, then it is irreducible in $k[x_1, ..., x_n][t]$ by Gauss's Lemma.

But in particular, if $gh=f'$, then by Gauss's Lemma, we can assume $g, h \in k[x_1, ..., x_n][t]$. Then we have $(ng)h=mf$. Each prime factor of $m$ must divide $ng$ or $h$, and so canceling prime factors, we end up with a factorization of $f$.

I haven't checked this too carefully. Does this seem right?

Answer 2

Your assertion is false already for $n=1$:

The polynomial $f(t)=x+xt\in k(x)[t]$ is irreducible in $k(x)[t]$, like all degree one polynomials over a field (which here is $k(x)$).
However it is reducible in $k[x][t]=k[x,t]$ since it factors as $f(t)=x+xt=x(1+t)$.
The apparent paradox is eexplained by the fact that $x$ is invertible in $k(x)[t]$ and so doesn't count as a factor of a factorization, whereas it is not invertible $k[x][t]$ and thus counts as a bona fide factor in a factorization.

Edit
This post answers the original question, namely:
"Let $f(t)$ be an irreducible polynomial in $k(x_1, \ldots, x_n)[t]$. Show that $f(t)$ is also irreducible in $k[x_1,\ldots,x_n][t]$. "

Answer 3

For clarity I'll post a new answer in order to explain Hartshorne's reasoning.

We have a monic irreducible polynomial $g(x_1,x_2,\cdots,x_n,Y)\in k(x_1,\cdots,x_n)[Y]$ with $g(x_1,x_2,\cdots,x_n,y)=0$.
We can write $g(x_1,x_2,\cdots,x_n,Y)=Y^s+a_1Y^{s-1}+\cdots a_s$ with $a_i=a_i(x_1,x_2,\cdots,x_n)\in k(x_1,\cdots,x_n)$.
Then we write $a_i=\frac {b_i}{d}$ where $b_i=b_i(x_1,x_2,\cdots,x_n),d=d(x_1,x_2,\cdots,x_n)\in k[x_1,\cdots,x_n]$ are polynomials.
In other words we have considered a common denominator $d$ for all our rational functions $a_i$.

The equation $g(x_1,x_2,\cdots,x_n,y)=y^s+a_1y^{s-1}+\cdots a_s=0$ after multiplication by $d$ then becomes : $$ dy^s+b_1y^{s-1}+\cdots b_s=0 $$ Dividing $d,b_1,\cdots b_s$ by their gcd $g$ in $k[x_1,\cdots,x_n]$, we get $\delta y^s+\beta _1 y^{s-1}+\cdots \beta _s=0$, with $\delta=\frac {d}{g}, \beta_i=\frac {b_i}{g}\in k[x_1,\cdots,x_n]$.
Hartshorne's polynomial is then $f(x_1,\cdots,x_n,Y)=\delta Y^s+\beta _1 Y^{s-1}+\cdots \beta _s\in k[x_1,\cdots,x_n][Y]$.

This polynomial $ f(x_1,\cdots,x_n,Y)$ is indeed irreducible in $k[x_1,\cdots,x_n][Y]$ by the criterion in your course note since it is primitive in $k[x_1,\cdots,x_n][Y]$ and irreducible in $k(x_1,\cdots,x_n)[Y]$.
(Recall that $\frac {\beta_i}{\delta}=\frac {b_i}{d}=a_i$ , so that $f=g \delta$ is indeed irreducible in $k(x_1,\cdots,x_n)[Y]$)