In $\mathbb Z[x]$, find an ideal $P$ such that $R/P$ consists of $4$ elements.

Question

I'm working on this problem : In $Z[x]$, find an ideal $P$ such that $R/P$ consists of $4$ elements.

My idea is to try and construct a suitable quotient of $Z[x]$ and an appropriate ideal generated by an irreducible polynomial, so that $Z[x]/P$ is isomorphic to the field with $4$ elements $F_4$. However, I'm only familiar with the theory that $Z_p[x]/P$, where $P$ is the ideal generated by an irreducible polynomial of degree $n$, is isomorphic to the field with $p^n$ elements.

How can I carry this over to a similar type of quotient of $Z[x]$ by a single ideal? I thought to maybe take the quotient $Z[x]/4Z$ first to achieve $Z_4%$, then quotient this new ring by an ideal generated by a linear polynomial (degree $1$) to construct a field with $4^1 = 4$ elements. However, this is sort of a "double quotient" that makes it difficult for me to see what the original ideal $P$ that we quotient $Z[x]$ by would be.

Thanks!

Answer 1

Let $P := <2 > + <x^2+x+1>$ be the $\Bbb Z[x]$ ideal generated by the polynomial $x^2+x+1$ and the constant $2$. $$\frac {\Bbb Z[x]}{P}\approx\frac{\frac {\Bbb Z[x]}{<2>}}{<x^2+x+1 +<2>>}\approx\frac{\Bbb F_2[x]}{<x^2+x+1>}\approx\Bbb F_4$$

Answer 2

The easiest example is $P=(4,x)$. Then ${\Bbb Z[x]}/{P}\cong {\Bbb Z}_4$. I think this is the one you have in mind.

Another simple example is $P=(2,x^2)$. Then ${\Bbb Z[x]}/{P}\cong {\Bbb Z}_2[u]$, where $u^2=0$, has four elements: $0,1,u,1+u$.