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I have a ring $R$ and I want to prove (or disprove) that $R[x,y]/(x-y^2) \cong R[y]$.

My idea is to define a ring homomorphism $\phi$ such that $\phi$ is the identity on $R$ and such that $\phi(y) = y$. This means that $\phi(x) = \phi(y^2) = y^2$.

This is is an injective homomorphism because $\phi$ is the identity for polynomials in $y$ and every polynomial has a unique representation as a polynomial in $y$.

This is clearly surjective because $\phi^{-1}(p(y)) = p(y)$ and therefore it is an isomorphism.

I do not see any errors but I want to make sure that what I am writing makes sense.

Is there a more general statement involving polynomial rings? For example, here is my conjecture:

$$R[x_1, x_2, \ldots, x_n]/(p(x_1, x_2, \ldots, x_n)) \cong R[x_1, x_2,\ldots, x_{n-1}]$$ where $p$ is a polynomial in $x_1, \ldots, x_n$ such that there is a term of degree $1$ in the polynomial with invertible coefficient.

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  • $\begingroup$ Generally, you might find it useful to apply the Fundamental Homomorphism Theorem to the appropriate surjective homomorphism, here $\phi\colon R[x,y]\to R[y]$. $\endgroup$ – Ted Shifrin Dec 18 '14 at 17:17
  • $\begingroup$ @user26857 Thank you. I have modified my conjecture. $\endgroup$ – user157227 Dec 18 '14 at 17:22
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Hint. In general, $A[T]/(T-a)\simeq A$ for $a\in A$. (In your case consider $A=R[y]$ and $a=y^2$.)

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In general, to show that a $R/I$ is isomorphic to another ring $S$, it is easier to produce a surjective homomorphism $R \to S$ that has kernel $I$ rather than an explicit isomorphism $R/I \to S$.

In your case, consider the homomorphism

$$R[x,y]\to R[y]\\x\mapsto y^2\\y\mapsto y$$

Is it surjective? What is its kernel?

Note that $R[x,y] = (R[y])[x]$, so this is just the evaluation map $x \mapsto y^2$.

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Another hint: consider $\phi(f(x,y))=f(y^{2},y)$ with kernel $(x-y^2)$

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