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# Understand the problem

Find the number of ring homomorphism from $\Bbb Z[x,y]$ to $\Bbb F_2[x]/(x^3+x^2+x+1)$

##### Source of the problem
TIFR 2019 GS Part A, Problem 16
Algebra
Moderate
##### Suggested Book
Abstract Algebra, Dummit and Foote

Do you really need a hint? Try it first!

Now observe that $(x^3+x^2+x+1)=(x+1)^3$ in $\Bbb F_2[x]$. Then can you think about the ring $\Bbb F_2[x]/(x^3+x^2+x+1)=\Bbb F_2[x]/(x+1)^3$?

Then we have to find the number of ring homomorphism from $\Bbb Z[x,y]$ to $\Bbb F_2[x]/(x^3+x^2+x+1)=\Bbb F_2[x]/(x+1)^3$. How many elements does the ring have?

$\Bbb F_2[x]/(x^3+x^2+x+1)=\Bbb F_2[x]/(x+1)^3\neq\Bbb F_2$. It is not a field 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 is the next move?

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, $Bbb F_2[x]/(x+1)^3=\{a+bx+cx^2: a,b,c \in \Bbb F_2\}$ has $2^3$ elements.

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