Prove $\frac{\mathbb Z[X,Y]}{(5,X^{2}-Y,XY+X+1)}$ is a field 
Prove $\frac{\mathbb Z[X,Y]}{(5,X^{2}-Y,XY+X+1)}$ is a field.

I thought to prove that this is isomorphic with $\mathbb{\mathbb Z_{5}(X)}$, and because $5$ is prime it will follow that it's a field.
I wanted to use the first isomorphic theorem.
I wanted to use the map $\phi: Z[X,Y]\mapsto\mathbb{Z}_{5}(X)$, $f(x,y)\mapsto f(x,x^{2})$.
Now I'm proving that 1) $\phi$ is a morphism 2) $\phi$ is surjective 3)$\ker\phi=(y-x^{2},x^{3}+x+1,5)$

*

*take $x,y \in \mathbb{Z[X,Y]}$ random then:


*

*$\phi(x+y)$=$\phi(\sum((a_{i1}a_{i2}+b_{i1}b_{i2})X^{i1}Y^{i2})$=$(\sum((a_{i1}a_{i2}+b_{i1}b_{i2})X^{i1}Y^{i2})$=$\sum((a_{i1}a_{i2}X^{i1}Y^{i2})+\sum(b_{i1}b_{i2})X^{i1}Y^{i2})$=$\phi(x)$+$\phi(y)$

*$\phi(xy)$=$\phi(\sum((a_{i1}a_{i2}b_{i1}b_{i2})X^{i1}Y^{i2})$=$(\sum((a_{i1}a_{i2}b_{i1}b_{i2})X^{i1}Y^{i2})$=$\sum((a_{i1}a_{i2}X^{i1}Y^{i2})\sum(b_{i1}b_{i2})X^{i1}Y^{i2})$=$\phi(x)$$\phi(y)$


*i don't know how to prove this



*let's prove two inlcusions.


*

*first let $f\in ker\phi$ then $f\in (Z[X,Y])([Z])$. We use the division algorithm, then there exist an $q(x,y)$ and a $r(x,y)$ so that $f(x,y)$=$q(x,y)(x^{3}+x+1)$+$r(y-x^{2})$+$5$
met $deg(x)<deg(x^{3}+x+1)=3$
I'm not sure how to prove those things but this is what i already have. Can someone help me further.
EDIT: my answer that i tried to prove is wrong. Some of you write a solution down. But i still need to prove that it's isomorphic with your solution and i'm still struggeling with the same question how to do that exactly
EDIT:
So the people who answered my question (thank you for that) don't really see my problem now.
Well now after you guys helpt me I want to prove that
$\frac{Z[X,Y]}{5,X^{2}-Y,XY+X+1}$ is isomorphic with $\frac{F_{5}[X]}{(X^{3}+X+1)}$.
So I need to prove that for the map the map $\phi$:$Z[X,Y]$ $\mapsto$$\frac{F_{5}(X)}{X^{3}+X+1}$:$f(x,y)$$\mapsto$$f(x,x^{2})$.
Now I'm proving that 1) $\phi$ is a morphisme 2) $\phi$ is surjective 3)ker$\phi$=$(y-x^{2},x^{3}+x+1,5)$
I'm stuck with proving these three things correctly
 A: Hint:
$$
\frac{\mathbb Z[X,Y]}{\langle 5,X^{2}-Y,XY+X+1 \rangle}
\cong
\frac{\mathbb Z[X,X^2]}{\langle 5,0,X^3+X+1 \rangle}
\cong
\frac{\mathbb Z[X]}{\langle 5,X^3+X+1 \rangle}
\cong
\frac{\mathbb F_5[X]}{\langle X^3+X+1 \rangle}
$$
so it reduces to proving that $X^3+X+1$ is irreducible mod $5$, which is easy since the degree is $3$.
A: As the other answers tell you,
$$\frac{\mathbb Z[X,Y]}{\langle 5,X^2-Y,XY+X+1 \rangle}\cong\frac{\mathbb Z[X]}{\langle 5,X^3+X+1 \rangle}\cong\frac{\mathbb Z_5[X]}{\langle X^3+X+1 \rangle}.$$
One way to prove this is to write explicit isomorphisms. I will not do this, since even though it works, it hides how one would come up with this chain of isomorphisms.
What you need is two theorems:


*

*(Third Isomorphism Theorem) If $I,J$ are ideals of $R$ and $I\subseteq J\subseteq R$, then $\frac RJ \cong \frac{R/I}{J/I}$.

*Let $\varphi: R\to S$ be a ring isomorphism and $I$ ideal of $R$. Then $R/I \cong S/\varphi(I)$.


So, in the first isomorphism in the chain, first use the Third Isomorphism Theorem to get
$$ \frac{\mathbb Z[X,Y]}{\langle 5,X^2-Y,XY+X+1 \rangle} \cong
\frac{\mathbb Z[X,Y]/ \langle X^2 - Y \rangle}{\langle 5,X^2-Y,XY+X+1 \rangle / \langle X^2 - Y \rangle}.$$
Now we want isomorphism $\varphi\colon \frac{\mathbb Z[X,Y]}{\langle X^2 - Y \rangle}\to \mathbb Z[X]$. Since we want $X^2-Y$ to be sent to $0$, first look at the evaluation map
\begin{align}
\mathbb Z[X,Y] &\to \mathbb Z[X],\\
X &\mapsto X, \\
Y &\mapsto X^2,
\end{align}
which you can verify is surjective and has kernel $\langle X^2 - Y \rangle$. That defines $\varphi$ by the First Isomorphism Theorem.
Finally, we need to calculate $\varphi(\langle 5,X^2-Y,XY+X+1 \rangle / \langle X^2 - Y \rangle)$. Notice that
\begin{align}
5 + \langle X^2 - Y \rangle &\mapsto 5,\\
X^2 - Y + \langle X^2 - Y \rangle &\mapsto 0,\\
XY+X+1+\langle X^2 - Y \rangle &\mapsto X^3+X+1.
\end{align}
Therefore, $\varphi(\langle 5,X^2-Y,XY+X+1 \rangle / \langle X^2 - Y \rangle) = \langle 5, X^3 + X + 1\rangle$  and
$$\frac{\mathbb Z[X,Y]/ \langle X^2 - Y \rangle}{\langle 5,X^2-Y,XY+X+1 \rangle / \langle X^2 - Y \rangle}\cong \frac{\mathbb Z[X]}{\langle 5, X^3 + X + 1\rangle}.$$
I will leave the second isomorphism in the chain to you.
A: If $Y=X^{2}$, then $\frac{\mathbb{Z}[X,Y]}{5,X^{2}-Y,XY+X+1}=\frac{\mathbb{F}_{5}[X]}{X^{3}+X+1}$. It should be easy to see that $X^{3}+X+1$ is irreducible in $\mathbb{F}_{5}$.
