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Let $k$ be a field. Let $a_1,\cdots,a_n$ be algebraic over $k$. Then $k[a_1,\cdots,a_n]=k(a_1,\cdots,a_n)$. Let $f_i(a_1,\cdots,a_{i-1},X_i)$ be the minimal polynomial of $a_i$ over $k(a_1,\cdots,a_{i-1})$. Suppose that $R_{i}(X_1,\cdots,X_n) \in k[X_1,\cdots,X_n]$ such that $R_i(a_1,\cdots,a_i,X_{i+1},\cdots,X_n)=0$. Then why is it true that $f_i(a_1,\cdots,a_{i-1},X_i)$ divides $R_i(a_1,\cdots,a_{i-1},X_i,\cdots,X_n)$? What confuses me is that $R_i(a_1,\cdots,a_{i-1},X_i,\cdots,X_n)$ is a polynomial in many variables $X_i,\cdots,X_n$.

Motivation: Matsumura, Commutative Ring Theory, proof of Theorem 5.1 page 32.

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You can think of the polynomial $R_i(X_1,...,X_i,X_{i+1},...,X_n)$ as the sum of monomials in $X_{i+1},...,X_n$ with coefficients in $k[X_1,...,X_i]$. The only way $R_i(a_1,...,a_i,X_{i+1},...,X_n)$ can be 0 is if each coefficient is 0. But since each coefficient lies in $k[X_1,...,X_i]$, it can only go to 0 if it is divisible by $f_i$. Since each coefficient is divisible by 0, everything is.

As an example over the rationals, if $f(x,y)=g(x)y+h(x)y^2$, and if $f(\sqrt{3},y)=0$, then we must have $g(\sqrt{3})=h(\sqrt{3})=0$, so that $x^2-3$ divides $h(x)$ and $g(x)$, and thus $f(x)$.

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    I think there are a couple of typos, e.g. "each coefficient lies in $k[X_1,\cdots,X_i]$" should be "each coefficient lies in $k[a_1,\cdots,a_i]$".2012-12-26