double quotient ring isomorphic to polynomials ring

Question

This question is quite specific, so apologies in advance. I don't have a clue of how to approach this. Answers and links to more general versions of this will also be very welcomed.

Let $F$ be a field and $A = F[x,y,z]$. $J=\langle xy-z^2 \rangle$ is an ideal in $A$. Denote $R=A/J$, and let $I=\langle x+J, y+J \rangle =\langle x+J \rangle + \langle y+J \rangle $ be an ideal in $R$.

I need to prove that $R/I \cong F[w]$.

Answer 1

By basic commutative algebra there are canonical isomorphisms $$(A/\langle xy-z^2\rangle)/\langle x+J,y+J\rangle\cong A/\langle x,y,xy-z^2\rangle=A/\langle x,y,z^2\rangle\cong F[z]/\langle z^2\rangle,$$ where the middle identity comes frome the simple fact that $\langle x,y,xy-z^2\rangle=\langle x,y,z^2\rangle$. It is not hard to prove (exercise for you?) that for any field $F$ you have $$F[z]/\langle z^2\rangle\not\cong F[w].$$ In particular, what you are asked to prove is false.

Answer 2

Assuming you meant $I = \langle x + J, z + J\rangle $:

There’s a bijection between $I \subseteq A / J$ and $J \subseteq I’ \subseteq A$ given by $\pi:A \to A / J$, $x \mapsto x + J$, which gives $\pi (I’) = I$. We can verify that $\pi (xA + zA) = I$, that is

$$I = I’ / J = (xA + zA) / J$$

By the Third Isomorphism Theorem we have

$$R/I = A/J/I = (A/J)/(A/I’) \simeq A /I’ = A/(xA+zA)$$

Finally, we can construct an isomorphism (“cancel out x and z”) and see that $$A/(xA+zA) = F[x, y, z] / (xF[x, y, z] + zF[x, y, z]) \simeq F[w]$$

Thus $R/I \simeq F[w]$.

Answer 3

Let's try and clear up things before starting. If $\tilde{x}=x+J$, $\tilde{y}=y+J$ and $\tilde{z}=z+J$ as elements of $R=A/J$, then all three elements are nonzero (easy degree considerations); however, $\tilde{z}^2=0+J\in I=\langle\tilde{x},\tilde{y}\rangle$. Hence $I$ is not a prime ideal. Thus the request to prove that $R/I\cong F[w]$ should be interpreted in the sense

prove that $R/I$ is generated by a single element as an $F$-algebra.

It is not possible that $w$ is an indeterminate over $F$, because $R/I$ is not a domain.

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Let $A$ be a ring with an ideal $J$. If $R=A/J$ and $I$ is an ideal of $R$, there is a unique ideal $I'$ of $A$ such that $J\subseteq I'$ and $I=I'/J$.

Also, $R/I\cong A/I'$.

What's $I'$ in your case? It is simply $\langle x,y\rangle+J$ (prove it).

Therefore the ring $A/I'$ is generated, as an $F$-algebra, by a single element, namely the image of $z$.