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I know that if $f(x)$ is irreducible then $\langle f(x) \rangle$ is a prime ideal. Then I thought: is it maximal? Then I search about it, I find that it is not maximal ideal but cannot find any proof.

Any help is appreciated. Thanks.

Robert Soupe
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Pradip
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2 Answers2

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An ideal $I \subset \mathbb Z [X]$ is maximal if and only if the quotient $\mathbb Z [X] / I$ is a field.

Suppose $f \in \mathbb Z [X]$ is irreducible and let $I = (f)$ the ideal generated by $f$.

If $f$ is degree zero (constant) then $\mathbb Z[X] / (f) \simeq \mathbb Z/(f)[X]$. Do you see why this is not a field? (hint: find an inverse of $X$)

If $f$ is of nonzero degree then $\mathbb Z[X] / (f) \simeq \mathbb Z[\alpha]$ where $\alpha$ is a root of $f$. Do you see why this is not a field? (hint: find an inverse of 2 or 3 or any $n \in \mathbb Z$).

Ruben
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  • This is essentially the same as the accepted answer, but a little more complicated (the last isomorphism relies on Gauss Lemma). Proving that some integer is not invertible means that this is not a divisor of the leading coefficient of $f$. – user26857 Jun 19 '19 at 19:42
  • @user26857 I agree. The approach is only slightly different. – Ruben Jun 20 '19 at 12:56
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If $f$ has degree 0, i.e. $f$ is a constant polynomial with a prime value $p$, then $\langle p \rangle \subsetneqq \langle p, x \rangle \subsetneqq \langle 1 \rangle$.

If $f$ has positive degree, let $p$ be a prime number not dividing the leading coefficient of $f$. Then $\langle f \rangle \subsetneqq \langle f, p \rangle \subsetneqq \langle 1 \rangle$.

Daniel Schepler
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  • Just a comment: $I=\langle p,x\rangle$ is a maximal ideal in $\Bbb Z[x]$, because the quotient $\Bbb Z[x]/I$ is isomorphic to $\Bbb Z_p$. I think that ideals like $\langle p, x-a\rangle$ are all the maximal ideals. – ajotatxe Jun 18 '19 at 18:45
  • @Daniel Schepler I did not get your 2nd case. Do u claim $\langle f, p \rangle$ is maximal ideal in $\mathbb{Z}[x]$? But I think this is true when f is irreducible in $\mathbb{Z}_{p}[x] $. For example take f(x) =x^4-10x^2+1 then your 2nd case will fail. – Pradip Jun 18 '19 at 18:54
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    No, I don't claim $\langle f, p \rangle$ is a maximal ideal, just that $\langle f \rangle$ is not a maximal ideal because $\langle f, p \rangle$ is a bigger ideal which still isn't the unit ideal. – Daniel Schepler Jun 18 '19 at 19:04
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    @ajotatxe No, the general maximal ideal is $\langle p, f \rangle$ where $\bar f \in \mathbb{F}p[x]$ is irreducible. So for example, $\langle 7, x^2 + 1 \rangle$ is a maximal ideal (and has quotient field isomorphic to $\mathbb{F}{49}$). – Daniel Schepler Jun 18 '19 at 19:07
  • @Daniel Schepler Is it necessary to choose p to be a prime not dividing leading coefficient of f(x) in 2nd case ? If I choose p to be any prime then what is problem? – Pradip Jun 19 '19 at 06:02
  • For example, if $f(x) = 2x + 1$ and you choose $p = 2$ then $\langle 2x + 1, 2 \rangle = \langle 1 \rangle$. – Daniel Schepler Jun 19 '19 at 16:05