Is the ideal generated by $y$ and $x^2+yx+1$ in $Z[x, y]$ a prime ideal? 
Is the ideal generated by $y$ and $x^2+yx+1$ in $Z[x, y]$ a prime ideal?

I'm new to ring theory etc, so the following solution may be totally wrong. Please let me know if it is correct. Thanks.
$\langle y \rangle \subset \langle y, x^2+yx+1 \rangle$. By third isomorphism theorem $$\frac{Z[x, y]}{\langle y, x^2+yx+1 \rangle}\cong \frac{\frac{Z[x, y] }{\langle y \rangle } }{ \frac{ \langle y, x^2+yx+1 \rangle}{ \langle y \rangle}}\cong \frac{Z[x] }{ \langle x^2+1 \rangle}$$ which is an integral domain.
 A: Why is true the isomorphism $\cong^*$? $$ \frac{\frac{\mathbb{Z}[x, y] }{\langle y \rangle } }{ \frac{ \langle y, x^2+yx+1 \rangle}{ \langle y \rangle}}\cong\frac{\frac{\mathbb{Z}[x, y] }{\langle y \rangle } }{ \frac{ \langle y \rangle+\langle  x^2+1 \rangle}{ \langle y \rangle}}\cong^* \frac{\mathbb{Z}[x] }{ \langle x^2+1 \rangle}.
$$
I have a direct verification. Define $\phi :\mathbb{Z}[x, y] \to\frac{\mathbb{Z}[x] }{ \langle x^2+1 \rangle}$ by $f(x, y)\to f(x, 0)+\langle x^2+1 \rangle$. Clearly, $f$ is a subjective ring homomorphim. Now let $f\in\mathbb{Z}[x, y] $ such that $\phi(f)=0$. that is, $f(x, 0)\in\langle x^2+1 \rangle$. We can wright $f$ of the form $f=a_0+yh+xg$, where $h\in \mathbb{Z}[x, y]$, $g\in\mathbb{Z}[x]$, and $a_0\in \mathbb{Z}$. Since $f(x, 0)\in\langle x^2+1 \rangle$, we have $a_0+xg\in\langle x^2+1 \rangle$. Hence, $f\in \langle y, x^2+yx+1 \rangle$. Thus, $Ker(\phi)\subseteq \langle y, x^2+yx+1 \rangle$. Clearly, $\langle y, x^2+yx+1 \rangle \subseteq Ker(\phi)$ and so we have
$$\frac{Z[x, y]}{\langle y, x^2+yx+1 \rangle}\cong \frac{Z[x] }{ \langle x^2+1 \rangle}.$$
Now since $\frac{Z[x] }{ \langle x^2+1 \rangle}$ is an integral domain we are done.
