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I am trying to show that the quotient ring $F[X, Y]/(X-Y)\cong F[X]$, where $F$ is a field and $(X-Y)$ is the ideal generated by $X-Y$.

I have the idea to define a map $\phi: F[X, Y] \rightarrow F[X]: \phi(f(X,Y)) = f(X, X)$. If $\ker \phi = (X-Y)$ then the result follows from the first isomorphism theorem, but I'm having trouble proving this is the case.

Clearly $(X-Y) \subset \ker \phi $, but how can we show that $f(X, X)=0 \implies f(X, Y) = r(X, Y)(X-Y)$?

fredgeot
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  • You can use this: https://math.stackexchange.com/a/664570/732532 – morrowmh Jun 04 '22 at 16:24
  • Look at $f$ as a polynomial in $X$ with coefficients in $F[Y]$. Divide this polynomial by the (monic) linear polynomial $X-Y$: $f(X,Y)=r(X,Y)(X-Y)+s(Y)$. Set $X=Y$: $0=f(Y,Y)=s(Y)$. Thus $s=0$ and $f=r\cdot(X-Y)$. –  Jun 04 '22 at 16:24
  • Consider https://math.stackexchange.com/questions/3105440/on-the-kernel-of-the-map-kx-y-to-kt-x-mapsto-t-y-mapsto-t?noredirect=1 – Arthur Jun 04 '22 at 16:24
  • Ah thanks I've got it, I was just missing that the factor theorem holds for polynomials over a field. – fredgeot Jun 05 '22 at 08:48

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