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

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, $(Y-X^2)\supset\ker\phi$.