Find the field that $\mathbb{Z}_7[x,y]/\langle y-2x^2, 4xy + y +1 \rangle$ is isomorphic to See the title. Here the set $\mathbb{Z}_7 = \{0,1,2,3,4,5,6\}$ is considered a ring with the obvious operations $+$ and $\cdot$; and $I := \langle y-2x^2, 4xy + y + 1 \rangle$ is an ideal in $\mathbb{Z}_7[x,y]$.
My idea was to show in two steps that:
$$ \mathbb{Z}_7[x,y]/I \cong \mathbb{Z}_7[x] / J \cong \mathbb{F}_{7^3}, $$
where $J := \langle x^3+2x^2+1 \rangle$.
The first isomorphism is true (on an intuitive level) because $y + I = 2x^2 + I$, so that:
$$(4xy + y + 1) + I = (8x^3 + 2x^2 + 1) + I = (x^3 + 2x^2 + 1) + I $$
To see why the second isomorphism holds, we need a bit more work. We know that $\mathbb{Z}_7[x]/\langle x^3 + 2x^2 + 1 \rangle$ is a field, because $x^3 + 2x^2 + 1$ is irreducible over $\mathbb{Z}_7$ (since no $x \in \mathbb{Z}_7$ is a root and its degree is 3). Now, "division" by $\langle x^3 + 2x^2 + 1 \rangle$ effectively gets rid of every polynomial in $\mathbb{Z}_7[x]$ of degree 3 or higher. We can reduce a polynomial of degree $\geq$ 3 to degree $\leq$ 2 by iteratively using the relation $x^3 = -2x^2-1 = 5x^2 + 6$. Moreover, every polynomial in $\mathbb{Z}_7[x]$ of degree $\leq$ 2 has a distinct representation in $\mathbb{Z}_7[x]/\langle x^3 + 2x^2 + 1 \rangle$. This means that $|\mathbb{Z}_7[x]/\langle x^3 + 2x^2 + 1 \rangle| = 7^3$, and hence this field must be isomorphic to $\mathbb{F}_{7^3}$.
I would like some tips on how to approach the details. Especially for the first isomorphism: I wanted to use the first isomorphism theorem to prove this. To this end, I used the ringmorphism:
$$\varphi : \mathbb{Z}_7[x,y] \to \mathbb{Z}_7[x]/J : p(x,y) \mapsto p(x,2x^2) + J$$
Then, I showed that $\ker(\varphi) = I$ and $\varphi(\mathbb{Z}_7[x,y]) = \mathbb{Z}_7[x]/J$. The first isomorphism theorem then tells us that $\mathbb{Z}_7[x,y]/\ker(\varphi) \cong \varphi(\mathbb{Z}_7[x,y])$, which is the desired result. I worked it out and it seems to be OK, but going through the details of the proof was rather cumbersome and I figured there should be an easier way.
The second isomorphism I think is pretty much proved already with the rough sketch I made earlier, I just need to work it out more precisely.
As a final question: is there maybe a way to find an isomorphism in just one step?
 A: A very useful trick to remember is that a multivariate polynomial ring is (isomorphic to) an iterated construction of polynomial rings. Namely, in your example, we have
$$
\Bbb F_7[x,y] \simeq \Bbb F_7[x][y].
$$
This is useful because polyomial rings in one variable may be more tractable.
For example: you can show that in general, if $T-\alpha \in A[T]$ is a degree $1$ monic polynomial in a polynomial ring $A$ and $P \in A[T]$ then there always exist unique $Q \in A[T]$ and $R \in A$ such that $P = (T-\alpha)Q+R$.
In particular, for any poynomial $p$ in $\Bbb F_7[x,y]$ there are unique $q \in \Bbb F_7[x,y]$ and $r \in \Bbb F_7[x]$ such that
$$
p(x,y) = (y-2x^2)q(x,y)+r(x).
$$
From this observation you can see that the map $\tau \colon \Bbb F_7[x,y] \to \Bbb F_7[x]$ sending $x \mapsto x, y \mapsto 2x^2$ has kernel precisely $(y-2x^2)$. Thus
$$
\Bbb F_7[x,y]/(y-2x^2) \simeq \Bbb F_7[x]
$$
and by the third isomorphism theorem we get
$$
\Bbb F_7[x,y]/I \simeq \Bbb F_7[x]/J.
$$
Now, you already observe that $J = (x^3+2x^2+1)$ is maximal, hence the latter quotient a field.
Since $\{[1],[x],[x^2]\}$ is a $\Bbb F_7$-basis for $\Bbb F_7[x]/J$ as a vector space, we have $\dim_{\Bbb F_7}\Bbb F_7[x]/J = 3$ and so $|\Bbb F_7[x]/J| = 7^3$. But then by uniqueness of finite fields, it has to be $\Bbb F_7[x]/J \simeq \Bbb F_{7^3}$.
Edit: here are some generalizations that use the exact same arguments: if $p = y-\alpha \in R[x][y]$ is of degree one, then $R[x,y]/(p) \simeq R[x]$ for any ring $R$, the isomorphism sends $x$ to $x$ and $y$ to $\alpha$.
Moreover, if $g \in R[x,y]$ then its image via this iso is $g(x,\alpha)$ and thus
$$
R[x,y]/(p,g) \simeq R[x]/(g(x,\alpha)).
$$
Fix $d = \deg g(x,\alpha(x))$. Then $\{[1],\ldots, [x^{d-1}]\}$ is a basis of $R[x]/(g(x,\alpha))$ as a free module.
In particular if $R = k$ is a field, then $k[X]/(g(x,\alpha))$ is a $d$-dimensional $k$-algebra, and if $g(x,\alpha)$ is irreducible then it is a field extension of dimension $d$. In that case, some computations follow:

*

*If $k = \Bbb F_q$ is finite, then $\Bbb F_q[X]/(g(x,\alpha)) \simeq \Bbb F_{q^d}$.


*If $d = 2 \neq \mathbf{char} (k)$, then $k[X]/(g(x,\alpha)) = k(\sqrt{d})$ with $d^2 \in k, d \not \in k$. Indeed, if $g(x,\alpha) = x^2+bc+c$ we have $k[X]/(g(x,\alpha)) \simeq k(\omega)$ with $\omega$ a root of $g(x,\alpha)$ in some algebraic closure, and we can apply the classification of quadratic extensions.
In general, if $R$ at least has the invariant basis property, you can (maybe) distinguish such quotients by looking at the total degree of a polynomial.
