Let $F$ be integer domain but not be field, then every ideal of $F[x]$ is prime ideal but not maximal?.

Now, I set a homomorphism $$\varphi: R[x]\to R$$ defined by $\varphi(a_0+a_1 x+\ldots+a_n x^n)=a_0$

then by First isomorphism theorem: $R[x]/\ker{\varphi}\cong R$

and $\ker{\varphi}=a_0+a_1 x+\ldots +a_n x^n$ such that $\varphi(a_0+a_1 x+\ldots+a_n x^n)=0$, or $a_0=0$, so $\ker{\varphi}=\langle x \rangle$. Because $R$ is integer domain, $R[x]/\langle x \rangle$ also is integer domain. That is $\langle x \rangle$ is prime ideal.

But why this is not maximal?.

In particular $R=\mathbb{Z}$, then $\langle x \rangle$ is not maximal because $\langle x \rangle \subset \langle 2,x \rangle$.

Is in general we also have $\langle x \rangle \subset \langle 2,x \rangle$?.


I think your claim is not true, because every non-zero ring must contain at least one maximal ideal and this follows by applying Zorn's lemma to the set of proper ideals partially ordered by natural inclusion.

  • $\begingroup$ Assuming its a ring with a unit $\endgroup$ – Belgi Dec 3 '12 at 8:16

For a domain $R$, $R[x]/(x) \cong R$, so unless $R$ is a field, $(x)$ cannot be maximal. But, may be I am misunderstanding what you are really trying to ask.

  • $\begingroup$ That is exactly thing I want ask. If $R$ isn't a field then $(x)$ isn's maximal ideal. Can you tell why? $\endgroup$ – Muniain Dec 3 '12 at 9:09

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.