How many elements does this ring have?

Question

I know that the following ring is not a field because the defining polynomial is reducible into two polynomials that are irreducible: $$\mathbb Z_2[X]/(x^5+x+1)$$ where $$x^5 + x + 1 = (x^2 + x + 1) (x^3 + x^2 + 1)$$

Is it possible to determine the number of elements of the ring?

Answer 1

$\mathbb{Z_2[x]}/(x^5+x+1) \cong \mathbb{Z_2[x]}/x^2+x+1 \times \mathbb{Z_2[x]}/x^3+x^2+1$ (Chinese Remainder Theorem)

$\mathbb{Z_2[x]}/(x^2+x+1)=\{\bar{1},\bar{0},\bar{x},\overline{(x+1)}$}

$\mathbb{Z_2[x]}/(x^3+x^2+1)=\{\bar{1},\bar{0},\bar{x},\overline{(x+1)}, \bar{x^2}, \overline {x^2+1}, \overline{x^2+x+1}, \overline{x^2+x}\}$

Answer 2

Factor the polynomial into irreducibles. I don't think your factorization is correct. Then use Chinese Remainder Theorem to express the ring as a product of fields.

Answer 3

If $F$ is a field and $f(x)\in F[x]$ a polynomial then $F[x]/f(x)$ has basis $\{1,x,\cdots,x^{\deg f-1}\}$ as a vector space over $F$. See if you can prove this! Thus it has cardinality $|F|^{\deg f}$ (for any $F$).

You don't have to do any factoring unless you want to determine the isomorphism type of the quotient, which is more information than its cardinality. One may factor $f$ and use Sun-Z (better known as CRT). If $f(x)$ is squarefree this will yield a product of extension fields of $F$.