Let \begin{align*} F = \frac{\mathbb{Z}[x,y]}{(5,x^2 - y, xy + x + 1)}. \end{align*} Prove that $F$ is a field. With what field is $F$ isomorphic?
I let the ideal $ I = (5, x^2 - y, xy + x + 1)$. I see that $x^2 - y + I = 0 + I$, so that $x^2 + I = y + I$. So working in this quotient ring, I can substitute the indeterminate $y$ for $x^2$. Also I have $5 + I = 0 + I$. So the coefficients will go only up to $5$. I can also see that $$ F \cong \frac{Z_5[x,y]}{(x^2 - y, xy + x + 1)}. $$ But then I still don't see with what this is isomorphic. I tried defining the map $$ \phi: \mathbb{Z}_5[x,y] \to \mathbb{Z}_5[x]: f(x,y) \mapsto f(x, x^2)$$ but the kernel of this does not equal $(x^2 - y, xy + x + 1)$.
$\frac {\mathbb Z[x,y]}{(5,x^2-y,xy+x+1)}\cong \frac {\mathbb Z[x,x^2]}{(5,x^3+x+1)}= \frac {\mathbb Z[x]}{(5,x^3+x+1)} $ we have $x^3+x+1$ irreducible polynomial in $\mathbb Z_5$ then $(5,x^3+x+1) $ is maximal ideal in $\mathbb Z[x]$ then $F$ is field.
or $\frac {\mathbb Z[x,y]}{(5,x^2-y,xy+x+1)}\cong \frac {\mathbb Z_5[x,y]}{(x^2-y,xy+x+1)}\cong\frac {\mathbb Z_5[x]}{(x^3+x+1)} $ , We have $\mathbb Z_5$ is a field then $\mathbb Z_5[x] $ is PID and $x^3+x+1$ irreducible polynomial in $\mathbb Z_5[x]$ then $(x^3+x+1) $ is maximal ideal in $\mathbb Z_5[x]$ then $F$ is field.