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I am trying to prove the following statement:

Let $k$ be an arbitrary field. Let $f(x, y) \in k[x, y]$ and $\beta \in k$. If $f(x, \beta) = 0 \in k[x]$, then $f(x, y)$ is divisible by $(y - \beta)$.

Here is my attempt:

Consider $f$ as an element of the polynomial ring $k(x)[y]$. Using division algorithm, we can write $f = (y - \beta) q + r$, where $q, r \in k(x)[y]$, and $r$ has degree [in $y$] strictly smaller than $1$, hence $r \in k(x)$. Since we are given $f(x, \beta) = 0 \in k[x] \subseteq k(x)$, we have $$0 = f(x, \beta) = (\beta - \beta) q(\beta) + r,$$ so $r = 0 \in k(x)$, and we just have $f = (y - \beta) q$. Now write $q$ explicitly as $q(y) = a_0(x) + a_1(x) y + \dots + a_n(x) y^n$ [with all $a_i(x)$'s in the field $k(x)$], and without loss of generality assume $a_n(x) \neq 0$. Let $b(x) \in k[x]$ be a common denominator for $a_0(x), a_1(x), \dots, a_n(x)$. Then $$b(x) f(x, y) = (y - \beta) [b(x) q(x, y)]$$ and note that $b(x) q(x, y) \in k[x, y]$. Since $(y - \beta) \in k[x, y]$ is irreducible, unique factorization implies that $(y - \beta)$ has to divide either $b(x)$ or $f(x, y)$. It cannot divide $b(x)$ because $b(x)$ has no $y$ term. Therefore $(y - \beta) \mid f(x, y)$.

Is my answer correct? Is there perhaps a more elementary method?

This is not a homework question. I was reading the proof of the very first lemma in Shafarevich's Basic Algebraic Geometry 1, and the second last sentence in that proof said "... $f(x, \beta)$ is not identically $0$, since otherwise $f(x, y)$ would be divisible by $y - \beta$ ..." It wasn't immediately obvious to me, and I'm just trying to fill in that gap.

nowhere
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  • Since $f(x,\beta)=0$ for all $x\in k$. I would assume that $k$ is not a finite field or you will be in trouble. If it does not divide, you will divide by $y-\beta$. The remainder is purely a function of $x$ as a finite degree polynomial. This is impossible by finite number of roots. – user45765 Jun 17 '17 at 03:12
  • @user45765 what are you saying? The assumption isn't that $f(x,\beta) = 0$ for all $x\in k$, the assumption is that $f(x,\beta)$ (which is an element of $k[x]$) is the zero polynomial, bypassing any issues that arise when one tries to equate a polynomial $p\in k[x]$ and the function $k\ni\alpha\mapsto p(\alpha)\in k$ it determines. – Stahl Jun 17 '17 at 03:29
  • @Stahl Yes. You read the statement correctly. I deliberately wrote "$f(x, \beta) = 0 \in k[x]$" to mean it's the zero formal polynomial. The field $k$ is completely arbitrary. Thanks for pointing that out. – nowhere Jun 17 '17 at 03:34
  • @Stahl Can I think $f(x,\beta):k\to k$ by plugging $k$ values? – user45765 Jun 17 '17 at 03:34
  • @user45765 You can, but it's a different beast (as you've noted). Or perhaps more precisely, every polynomial $p\in k[x]$ induces a function $ev_p : \alpha\mapsto p(\alpha)$, but the map $p\mapsto ev_p$ is not injective in general (but it seems you already know this?). I was asking why you brought this up because in the problem above, the equality $f(x,\beta) = 0$ is demanded as polynomials (not functions $k\to k$), removing the issue. – Stahl Jun 17 '17 at 03:37

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Your proof looks fine to me! Your method is already pretty elementary: it seems you only use the division algorithm for $K[x]$ where $K$ is a field, and the fact that $k[x,y]$ is a UFD when $k$ is a field, neither of which requires heavy machinery to prove.

Stahl
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