Is the Ideal $I = (x^2+y^2+x+y, x^2+y^2+xy)$ of $\mathbb Z[x,y]$ prime?
In $Z[x,y]/I$, we have the relation $x+y = -(x^2+y^2) = xy.$ Hence, we get that $(xy)^2 = (x+y)^2 = x^2+y^2+2xy = -(xy)+2xy = xy$, so $$xy(xy-1) = (x+y)(x+y-1) = 0.$$
By contradiction, suppose that $I$ is prime so $\mathbb Z[x,y]/I$ is an integral domain. Then, $xy(xy-1) = 0$ implies that $xy = 0 $ or $xy=1$. If $xy = x+y = 0$, then $0 = (x+y)y = xy+y^2 = y^2$. Similarly, we see that $x^2 = 0$. Thus, $x = y = 0$ in the quotient, meaning that $x,y \in I$. As a consequence, we obtain $\mathbb Z[x,y]/I \cong \mathbb Z$. However, so far I don't see how this can lead to a contradiction here.
On the other hand, if we had $xy=1$ in $\mathbb Z[x,y]/I$, of course one cannot have $x=0 $ or $y=0$, since then $1=0$ would imply that the ideal $I$ is not proper ($I$ is proper, because every polynomial in that ideal gives zero when evaluated at $(0,0)$). Now, from $xy=1$ we would also get that $x = x^2y$, so $x(y-x) = 0$ implies that $y=x$. Hence, we also get that $x^2 = y^2 = 1$. I believe that would allow me to conclude that $\mathbb Z[x,y]/I$ is spanned by $\{1,x,y\}$. However, I also don't see how to get a contradiction here.
How to proceed with this problem? I really believe it should not be a prime ideal, but so far I found no contradictions. Also, I noticed that $x^2+y^2+xy$ is an irreducible (so, prime) element in $\mathbb Z[x,y]$, but I am not sure how to use it here.