Number of Ring homomorphism form
$$\displaystyle\mathbb{Z[x,y]}\;\;\;\;\; to\;\;\;\; \frac{\mathbb{F_2[x]}}{\langle x^3+x^2+x+1 \rangle}$$
As we can see that $\displaystyle \frac{\mathbb{F_2[x]}}{\langle x^3+x^2+x+1 \rangle}$ is not a field because the ideal $\langle x^3+x^2+x+1 \rangle $ is not a irreducible element in $\mathbb{F_2[x]}$.Does i have to calculate the ideals of $\mathbb{Z[x,y]}$?,i have no idea how to proccede please help.
Thankyou
Adding to the comment thread above, here is a justification for why the answer is $2^6$:
There are $8$ elements in $\Bbb F_2[x]/(x^3+x^2+x+1)$. They are (represented by) all polynomials of degree $2$ or less: $$ 0\\ 1\\ x\\ x+1\\ x^2\\ x^2+1\\ x^2+x\\ x^2+x+1 $$ Now, we are looking for functions from the generating set $\{1, x, y\}\subseteq \Bbb Z[x, y]$ to $\Bbb F_2[x]/(x^3+x^2+x+1)$ that will give rise to homomorphisms. It turns out that any function that sends $1\mapsto 1$ will give rise to a homomorphsim (this is a property of $\Bbb Z[x, y]$, and polynomial rings in general: the $x$ and $y$ do not interact more than just the bare minimum for a ring, so they may be sent wherever, independently, without any issues). Which means we are entirely free to choose where to send $x$ and where to send $y$.
There are $8$ choices for where such a function can send $x$, and $8$ choices for where such a function can send $y$. Which is to say, there are $8^2 = 64$ choices in total for where to send $x$ and $y$. Each such choice, coupled with $1\mapsto 1$ will give rise to a homomorphism, no two such choices will give rise to the same homomorphism, and any homomorphism stems from such a choice. Thus there are $64$ homomorhisms in total.