Dtyjlui

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 $R/I cong F[w]$.

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

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

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 $R/I simeq F[w]$.

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, $tilde{z}^2=0+Jin I=langletilde{x},tilde{y}rangle$. Hence $I$ is not a prime ideal. Thus the request to prove that $R/Icong F[w]$ 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.

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 $Jsubseteq I'$ and $I=I'/J$.

Also, $R/Icong A/I'$.

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

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