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

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!

• What is $R$ here? – mathcounterexamples.net Oct 14 '19 at 19:58
• Hint: Consider an ideal $P$ with two generators. – Jyrki Lahtonen Oct 14 '19 at 20:00
• A useful generalization. I might even call this a duplicate of that, but I'm too tired to argue the case. – Jyrki Lahtonen Oct 14 '19 at 20:03

Let $$P := <2 > + $$ be the $$\Bbb Z[x]$$ ideal generated by the polynomial $$x^2+x+1$$ and the constant $$2$$. $$\frac {\Bbb Z[x]}{P}\approx\frac{\frac {\Bbb Z[x]}{<2>}}{>}\approx\frac{\Bbb F_2[x]}{}\approx\Bbb F_4$$
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: $$0,1,u,1+u$$.