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I'm trying to prove (from a previous post) that if $A=k[x,y,z]$ and $I=(x,y)(x,z)$ then $$\dfrac{(x,y)/I}{(x,yz)/I} \cong\dfrac{A}{(x,z)}.$$

I did this by defining the homomorphism $\phi: A \to ((x,y)/I)/((x,yz)/I)$ by $f(x,y,z) \mapsto \dfrac {f(x,y,0)+ I}{(x,yz)/I}$ with the extra property that $\phi$ kills all the constants in $f$ (not only all $z$). Then since it is surjective and its kernel equals $(x,z)$ I am done...Does this seem correct?

Second, I am trying to prove that $(x,yz)/I \cong A/(x,y,z)$. How can I do this?

I forgot to mention that we consider the $A$-module $M=A/I$, and that we want to prove that submodules are isomorphic.

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$\dfrac{(x,y)/I}{(x,yz)/I} \simeq\dfrac{(x,y)}{(x,yz)}=\dfrac{(x,y)}{(x,y)\cap (x,z)}\simeq\dfrac{(x,y)+(x,z)}{(x,z)}=\dfrac{(x,y,z)}{(x,z)}=\dfrac{(x,z)+(y)}{(x,z)}\simeq$ $\dfrac{(y)}{(x,z)\cap (y)}=\dfrac{(y)}{(xy,zy)}\simeq\dfrac{A}{(x,z)}$.

$\dfrac{(x,yz)}{I}=\dfrac{(x,yz)}{(x^2,xy,xz,yz)}\simeq\dfrac{(x)}{(x^2,xy,xz)}\simeq\dfrac{A}{(x,y,z)}$.

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