# Multivariate polynomial with univariate factor

Suppose $f \in \mathbb{Q}[x_1,x_2,\dotsc,x_i]$. How do I prove the following:

There exists $x_1 \in \mathbb{C}$ such that for all $x_2,\dotsc,x_i \in \mathbb{C}$, $$f(x_1,x_2,\dots,x_i) = 0$$ if and only if $f$ is divisible by some $g \in \mathbb{Q}[x_1]$.

(I suspect this is simply down to me not knowing proper definitions...)

## 1 Answer

In general, for any given constant $c\in{\mathbb C}$, an arbitrary polynomial $h$ can be Taylor-decomposed with respect to $x_1$ as

$$f(x_1,x_2,x_3,\ldots ,x_i)=\sum_{k=0}^d h_k(x_2,x_3,\ldots ,x_i) (x_1-c)^k$$

It is easy then to see that $f(c,x_2,x_3,\ldots, x_i)=0$ for any $x_2,x_3,\ldots, x_i$ iff $x_1-c$ divides $f$.

If $c$ is transcendental, then $f=0$ (exercise : why?). If $c$ is algebraic and $Q$ is the minimal polynomial of $c$ over $\mathbb Q$, show that $Q(x_1)$ divides $f$ iff $x_1-c$ divides $f$.

UPDATE : Suppose that $c$ is transcendental. Decompose $f$ as

$$f=\sum_{k=0}^d g_k(x_2,x_3,\ldots ,x_i) x_1^k$$

where $g_k$ are polynomials with rational coefficients, not involving $x_1$. Since $c$ is transcendental, the numbers $1,c,c^2,c^3, \ldots, c^d$ are linearly indepedent over $\mathbb Q$. So the hypothesis $f(c,\ldots)=0$ entails that all the $g_k$ are zero, so $f=0$.

Suppose that $c$ is algebraic, with minimal polynomail $Q$ (over $\mathbb Q$). Denote by $c_1=c,c_2, \ldots, c_r$ the conjugates of $c$ (over $\mathbb Q$). Since $x_1-c$ divides $f$, by conjugation $x_1-c_j$ also divides $f$, for any $j$. So $Q$ is divsible by the product of the $x_1-c_j$, which is none other than $Q(x_1)$.

• Also, wrt your last line - when $k=0$, $(x_1-c)^k$ is $1$ - is it okay to say that $x_1-c$ divides a function that doesn't use $x_1$? – user98041 Oct 1 '13 at 16:14
• @user98041 the hypothesis $f(c,\ldots)=0$ is equivalent to $h_0=0$, so the term for $k=0$ vanishes. – Ewan Delanoy Oct 1 '13 at 16:45
• @user98041 It appears one needs some Galois theory to answer your question. Do you know something about it ? – Ewan Delanoy Oct 1 '13 at 16:46
• @user98041 : try to work out the additional hints I inserted in my answer. I’ll elaborate according to your needs. By the way : no, this is not just a “know-the-definitions” exercise (although it is easy by MSE’s standards, which explains why it received no upvotes so far). – Ewan Delanoy Oct 1 '13 at 16:59
• @user98041 See my update. – Ewan Delanoy Oct 1 '13 at 19:13