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Find the number of ring homomorphism from $\mathbb Z[x,y]\to \mathbb{F}_2[x]/(1+x+x^2+x^3)$.

My attempt: the ring $Z[x,y]$ has three generators $1,x \ and\ y$ we want $1$ to map to $1.$ Since the ring $\mathbb{F}_2[x]/(1+x+x^2+x^3)$ has 8 elements we have total of $8\times 8$ ring homomorphisms.

Am I right ?

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It's fine, except that $1$ is not a generator of $\Bbb Z[x,y]$, only $x$ and $y$ are.
It's true, however, that the images of $x,y$ under a homomorphism $f:\Bbb Z[x,y] \to R$ can be arbitrary elements of $R$ and they uniquely determine $f$.

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A ring homomorphism $\varphi\colon R[x,y]\to S$ (where $R$ and $S$ are any commutative rings) is completely determined by assigning

  1. a ring homomorphism $\varphi_0\colon R\to S$,
  2. $\varphi(x)$,
  3. $\varphi(y)$.

This is essentially the universal property of polynomial rings.

In your case, $R=\mathbb{Z}$ and there exists a unique ring homomorphism $\mathbb{Z}\to S$, so you just need to assign $\varphi(x)$ and $\varphi(y)$, which makes for $|\mathbb{F}_2/(1+x+x^2+x^3)|$.

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