Find the number of ring homomorphism from $\Bbb Z[x,y]$ to $\Bbb F_2[x]/(x^3+x^2+x+1)$.
Now I have observed that $(x^3+x^2+x+1)=(x+1)^3$ in $\Bbb F_2[x]$. Then $\Bbb F_2[x]/(x^3+x^2+x+1)=\Bbb F_2[x]/(x+1)^3=\Bbb F_2$. Am I right here?.
Then we have to find the number of ring homomorphism from $\Bbb Z[x,y]$ to $\Bbb F_2$. Can anyone help me with this problem?
Edit: After looking some comments I realized how fool I was. $\Bbb F_2[x]/(x^3+x^2+x+1)=\Bbb F_2[x]/(x+1)^3\neq\Bbb F_2$. It is not a field(thanks to all the commenter) but it has $8$ elements as $\Bbb F_2[x]/(x^3+x^2+x+1)=\Bbb F_2[x]/(x+1)^3=\{a+bx+cx^2: a,b,c \in \Bbb F_2\}$ Now what?
As it is said in the comments, for any commutative ring $R$ there is a bijection between the homomorphisms $\phi:\mathbb{Z}[x,y]\to R$ and the elements in $R^2$. To any $(a,b)\in R^2$, $\phi(x)=a$ and $\phi(y)=b$ determines the homomorphism.
Hence, if one counts the number of elements in $\mathbb{F}_2[x]/(x^3+x^2+x+1)$, the number of homomorphisms you want is just the square.
Now, $$\Bbb F_2[x]/(x^3+x^2+x+1)=\Bbb F_2[x]/(x+1)^3\cong \Bbb F_2[x]/x^3$$ where the last isomorphim sends $x$ to $x+1$. Now, the elements in $\Bbb F_2[x]/x^3$ can be written all as $a_0+a_1 \alpha+a_2\alpha^2$, where $\alpha$ denotes the class of $x$ in $\Bbb F_2[x]/x^3$, and $(a_0,a_1,a_2)\in \Bbb F_2^3$ can be any element. Since no two such elements can be equal, there are eight of them.