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I am trying to solve the following question:

``Let $F$ be a field and let $F[x, y]$ denote the ring of polynomials in the variable $x$ and $y$ with coefficients in $F$. Suppose $f(x,y)$ belongs to $F[x, y]$. Apply the Factor Theorem to the ring $F[x]$ to show that $f(x, x) = 0$ if and only if $(x-y)$ is a factor of $f(x, y)$. More generally, show that $y - g(x)$ divides $f(x, y)$ if and only if $f(x, g(x)) = 0$, for $g(x)$ in $F[x]$.''

If $(x-y)$ is a factor of $f(x, y)$ then obviously $f(x, x) = 0$. But how do I prove the converse? Is there a version of the Factor Theorem for $F[x, y]$ that I need to use?

quasi
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Omar Khan
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1 Answers1

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If $R$ is a commutative ring, then for the ring $R[x]$ we have the Factor Theorem:

    Let $f\in R[x]$ and let $r\in R$. Then in $R[x]$ we have $(x-r){\,\mid\,}f\;$if and only if $f(r)=0$.

Now suppose $F$ is a commutative ring and $f\in F[x,y]$ is such that $f(x,x)=0$.

Let $R=F[x]$ and let $g\in R[y]$ be defined as $g(y)=f(x,y)$. \begin{align*} \text{Then}\;\;& f(x,x)=0\\[4pt] \implies\;& g(x)=0 \\[4pt] \implies\;& (y-x){\,\mid\,}g\;\text{in $R[y]$}&&\text{$\Bigl($by the Factor Theorem for $R[y]\Bigr)$} \\[4pt] \implies\;& (x-y){\,\mid\,}g\;\text{in $R[y]$} \\[4pt] \implies\;& (x-y){\,\mid\,}f\;\text{in $F[x,y]$} \\[4pt] \end{align*} as was to be shown.

quasi
  • 58,772