$\mathbb{Q}[X,Y]/(X,Y^{2}-1)$ is this a maximal ideal or a prime ideal? $\mathbb{Q}[X,Y]/(X,Y^{2}-1)$ is this a maximal ideal or a prime ideal?
So far i got:
$\mathbb{Q}[X,Y]/(X,Y^{2}-1) = \left(\mathbb{Q}[Y]\right)X/(X,Y^{2}-1)$ is isomorphic to 
$\mathbb{Q}[Y]/(Y^{2}-1)$ = $\mathbb{Q}[Y]/(Y+1)(Y-1)$. Because we now have zero divisors this can't be a domain, so its cant be a prime/maximal ideal.
I don't trust my proof because i don't know why the second step would be valid.
Please help :)
Kees
 A: Let's reassure you that the step of your reasoning you were unsure about can be done rigorously.
It's definitely right that $\Bbb Q[X,Y]=(\Bbb Q[X])[Y]$.
Then looking at $\mathbb{Q}[X,Y]/(X,Y^{2}-1) = \left(\mathbb{Q}[Y]\right)[X]/(X,Y^{2}-1)$, the idea would be to apply an isomorphism theorem to say that:
$$
\left(\mathbb{Q}[Y]\right)[X]/(X,Y^{2}-1)\cong \frac{\mathbb{Q}[Y][X]/(X)}{(X,Y^{2}-1) /(X)}
$$
The top is obviously $\Bbb Q[y]$ under the evaluation map $X\to 0$, and the bottom is the ideal $(Y^2-1)$ in that ring, so you indeed have $\frac{\mathbb{Q}[Y]}{(Y^{2}-1) }$.
A: You are right. The polynomials $Y - 1$ and $Y + 1$ are both outside the ideal $(X, Y^2 - 1)$, but their product is contained in it, thus it cannot be a prime ideal. Since any maximal ideal is prime, it cannot be maximal either.
A: It can't be maximal, as it is properly contained in both of the (distinct) maximal ideals $(X,Y\pm 1)$.  
A: In the end you don't need the equality $Q[Y]/(Y^2−1) = Q[Y]/(Y+1)(Y−1) $. It is sufficient to say that $(Y-1)$ and $(y+1)$ are zero divisors, so this can't be a domain, and therefore can't be prime (or maximal).
