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
Topic
Algebra
Difficulty Level
Moderate
Suggested Book
Abstract Algebra, Dummit and Foote

Start with hints

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.  

So, the number of ring homomorphisms is 2^6.

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