Can this quick way of showing that $K/(Y-X^2)\cong K$ be turned into a valid argument?

I’ve been trying to show that

K[X,Y]/(Y-X^2)\cong K[X]

where $K$ is a field, $K[X]$ and $K[X,Y]$ are the obvious polynomial rings over the indeterminates $X$ and $Y$ and $(Y-X^2)$ is the ideal generated by the polynomial $Y-X^2$. Though I’m sure there’s a fairly easy way to find an explicit isomorphism between the two rings, the following argument jumped out at me:

If we substitute in a value for $X$ – $x$, say, then the ideal $(Y-x^2)$ is a maximal ideal of $K[Y]$. So the quotient $K[Y]/(Y-x^2)$ is a field; in fact, the homomorphism $K[Y]\to K:P(Y)\mapsto P(x^2)$ and is clearly surjective, so the quotient is isomorphic to $K$.

I’d like to be able to deduce from this that $K[X,Y]/(Y-X^2)\cong F[X]$, but I can’t see a nice way to do it. I know that the ‘substitution’ maps $P(X,Y)\mapsto P(x,Y)$ are homomorphisms, but I can’t see a nice way of pulling all these homomorphisms back to the polynomial ring in two variables.

Or maybe I’m completely wrong and there is no way to turn this into a valid argument. Can anyone help me?

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Thanks to Hurkyl’s clue, I was able to answer this question myself.

We consider the homomorphism $\phi:K[X,Y]\to K[X]:P(X,Y)\mapsto P(X,X^2)$. It is easy to verify that $\phi$ is surjective so if we can show that it has kernel $(Y-X^2)$, then we can conclude that $K[X,Y]/(Y-X^2)\cong K[X]$ by the first isomorphism theorem.

To show that $\ker\phi = (Y-X^2)$: first note that clearly $Y-X^2\in\ker\phi$, so $(Y-X^2)\subset\ker\phi$. On the other hand, if $P\in\ker\phi$, then $X^2$ is a root of $P$ considered as a polynomial in $Y$ with coefficients in $K[X]$. So $(Y-X^2)\vert P$, by the factor theorem. Therefore, $(Y-X^2)\supset\ker\phi$. $\Box$

More explicitly, suppose $P\in\ker\phi$. Writing $P(X,Y)=\sum_{i=0}^n a_i(X)Y^i$, where the $a_i$ are polynomials in $X$ with coefficients in $K$, we know that:

\phi(P)(X)=\sum_{i=0}^n a_i(X)(X^2)^i=0

Subtracting these equations gives:

P(X,Y)=P(X)-0=P(X,Y)-\phi(P)(X) = \sum_{i=0}^na_i(X)(Y^i-(X^2)^i)

Then, noting that $Y^i-(X^2)^i=(Y-X^2)(Y^{i-1}+Y^{i-2}X^2+\dots+(X^2)^{i-1}$, we see that $(Y-X^2)$ is a factor of $P(X,Y)$; therefore, $(Y-X^2)\supset\ker\phi$.