I am wondering if this application of the Third Isomorphism Theorem for Rings is valid. In the polynomial ring $\mathbb{Q}[x, y]$ if we mod out by the ideal $(x, y)$ in $\mathbb{Q}[x, y]$, we of course get $\mathbb{Q}$ as the resulting ring.
I've shown this with an explicit isomorphism, but I was wondering if there's a more sophisticated way to do it with previous results (I'm on chapter 9.1 in Dummit and Foote). Since $(y) \subseteq (x, y)$, it follows by the Third Isomorphism Theorem that
$$\mathbb{Q}[x, y]/(x,y) \cong (\mathbb{Q}[x, y] / (y))/((x, y) / (y)) $$
Then we see that $\mathbb{Q}[x, y] / (y) \cong \mathbb{Q}[x]$ and that $(x, y) / (y) \cong (x) / ((x) \cap (y))$, but I don't believe $(x) \cap (y)$ vanishes in this case, since they are ideals in $\mathbb{Q}[x, y]$. For example, $xy$ is indeed in this ideal. Am I thinking about this properly, or am I missing a detail or two?
Thank you for taking a look.
If you accept that $\mathbb{Q}[x] \cong \mathbb{Q}[x,y]/(y)$ via the natural map, then you can see that the ideal $(x)$ goes to $(x,y)/(y)$, i.e. the isomorphism gives you an isomorphism of the quotients that you want: $\mathbb{Q}[x]/(x) \cong (\mathbb{Q}[x,y]/(y))/((x,y)/(y))$.
Be careful with quotients. If $A \cong B, C \cong D$ and $C$ is an ideal of A, $D$ an ideal of B, it is not true that $A/C \cong B/D$. It is important that the isomorphisms are actually the same, i.e. $C \cong D$ by restricting the isomorphism between $A$ and $B$.