2
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ What is $R$ here? $\endgroup$ – mathcounterexamples.net Oct 14 '19 at 19:58
  • 4
    $\begingroup$ Hint: Consider an ideal $P$ with two generators. $\endgroup$ – Jyrki Lahtonen Oct 14 '19 at 20:00
  • 1
    $\begingroup$ A useful generalization. I might even call this a duplicate of that, but I'm too tired to argue the case. $\endgroup$ – Jyrki Lahtonen Oct 14 '19 at 20:03
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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: $0,1,u,1+u$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.