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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 $\langle x \rangle \subset \langle 2,x \rangle$?.

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Oh, so deplorably low an accept rate you have ! –  DonAntonio Dec 3 '12 at 11:59
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2 Answers 2

up vote 5 down vote accepted

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.

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Assuming its a ring with a unit –  Belgi Dec 3 '12 at 8:16
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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.

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