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Problem:

Let $F$ be a field. Prove that if $p(x)$ is irreducible, then $\langle p(x) \rangle$ is a maximal ideal of $F[x]$.

Attempt:

Let $d(x),a(x),p(x) \in F[x]$ and suppose $\text{gcd}[a(x), p(x)] = d(x)$. Then $d(x) \mid p(x)$. So $p(x) = d(x)c(x)$ for some $c(x) \in F[x]$. Because $p(x)$ is irreducible, either $d(x)$ or $c(x)$ is a constant.

If $d(x)$ is a nonzero constant in $F$, then $\langle p(x) \rangle = F[x]$ by [previous proof]* and $\langle p(x) \rangle$ is not maximal because it generates all of $F[x]$.

If $c(x)$ is a nonzero constant, then: \begin{align*} p(x) &= d(x)c(x) \\ p(x) &= d(x)c && \text{$c(x) = c$ is a constant}\\ p(x)c^{-1} &= d(x)cc^{-1} \\ p(x)c^{-1} &= d(x) \\ d(x) &= p(x)c^{-1} \end{align*}

Since $d(x) \mid a(x)$, we have that $a(x) = d(x)e(x)$ for some $e(x) \in F[x]$. Then: \begin{align*} a(x) &= d(x)e(x) \\ a(x) &= p(x)c^{-1}e(x) && \text{substitution}\\ a(x) &= p(x)e(x)c^{-1} \end{align*}

This implies that $a(x)$ is a multiple of $p(x)$.

Questions:

My understanding is that if $d(x)$ is a nonzero constant, then $\langle p(x) \rangle$ cannot be maximal because it generates the entire ring. But it also seems to me that $c(x)$ being a constant presents a contradiction.

Does $a(x)$ being a multiple of $p(x)$ present a contradiction, because $a(x)$ and $p(x)$ are relatively prime?

If not, how can I use that $c(x)$ being constant shows that $\langle p(x) \rangle$ is maximal?

Second, guided attempt:

Let $J$ be an ideal containing $\langle p(x) \rangle$ that isn't equal to $\langle p(x) \rangle$. Then there must exist a polynomial $a(x)$ that is in $J$ but not in $\langle p(x) \rangle$. We'll show that $J = F[x]$ using that $p(x)$ and $a(x)$ are relatively prime.

Let $d(x),a(x),p(x) \in F[x]$ and suppose $\text{gcd}[a(x), p(x)] = d(x)$. Then $d(x) \mid p(x)$. So $p(x) = d(x)c(x)$ for some $c(x) \in F[x]$. Because $p(x)$ is irreducible, either $d(x)$ or $c(x)$ is a constant.

By [previous proof]*, if $d(x)$ is a nonzero constant in $F$, then $J=F[x]$ and we are done.

If, on the other hand, $c(x)$ is a nonzero constant, then: \begin{align*} p(x) &= d(x)c(x) \\ p(x) &= d(x)c && \text{$c(x) = c$ is a constant}\\ p(x)c^{-1} &= d(x)cc^{-1} \\ p(x)c^{-1} &= d(x) \\ d(x) &= p(x)c^{-1} \end{align*}

Since $d(x) \mid a(x)$, we have that $a(x) = d(x)e(x)$ for some $e(x) \in F[x]$. Then: \begin{align*} a(x) &= d(x)e(x) \\ a(x) &= p(x)c^{-1}e(x) && \text{substitution}\\ a(x) &= p(x)e(x)c^{-1} \end{align*}

This implies that $a(x)$ is a multiple of $p(x)$, which is a contradiction, because $a(x)$ and $p(x)$ are relatively prime. Thus, $\langle p(x) \rangle$ is a maximal ideal of $F[x]$.

* Previous proof: If $J$ is an ideal of $A$ and $J$ contains an invertible element $a$ of $A$, then $J = A$.

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  • $\begingroup$ $(d) = (a,p) = (1)\,$ or $(p)$ by $(p)$ irred, so $\,a\not\in (p)\Rightarrow\, (a,p)=(1)\,$ so $(p)$ is maximal. $\endgroup$ – Bill Dubuque May 13 '19 at 15:58
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    $\begingroup$ Hint $\ $ For principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e. $(a)\supseteq (b)\iff a\mid b,\,$ thus having no proper containing ideal (maximal) is the same as having no proper divisor (irreducible), $ $ i.e. $\\ \qquad\quad\begin{eqnarray} \\ (p)\,\text{ is maximal} &\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (d)\\ &\iff&\ p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ d\\ &\iff&\ p\ \ \text{ is irreducible}\\ \end{eqnarray}\ \ \ $ $\endgroup$ – Bill Dubuque May 13 '19 at 16:01
  • $\begingroup$ What is $a(x)$ in your proof? And how are you using it in the proof? $\endgroup$ – Julian Mejia May 13 '19 at 16:03
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    $\begingroup$ So, after reading the answer of Ehsaan, apparently your assumption was $(p(x))\subset I= (a(x))$ and you wanted to prove that $I=(p(x))$or $F[x]$? Is that right? This is what I was asking for, because in your proof, $(a(x))$ is not just a random polynomial. $\endgroup$ – Julian Mejia May 13 '19 at 16:58
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    $\begingroup$ So the proof boils down to the fact that irred $\,p\nmid a\,\Rightarrow\, (p,a) = 1\,$ where we can read the pair notation as either a gcd or ideal. This connection between gcds and principal ideals in PIDs allows us to transfer divisibility intuition from $\Bbb Z$ to arbitrary PIDs, e.g. see generalizations of Euclid' Lemma $\endgroup$ – Bill Dubuque May 13 '19 at 17:39
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Every ideal of $F[x]$ is principal. If $\langle p(x)\rangle$ is not maximal, then $p(x)$ must have a non-constant factor with a lower degree, which is impossible.

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Here is the proof, maybe that will settle your questions?

If $(p(x))$ is not maximal, then there is an ideal $I$ containing it. We'll show $I=F[x]$ or $I=(p(x)).

Since $F[x]$ is a principal ideal domain, we know $I=(f(x))$ for some polynomial $f(x)$. (This is probably where your argument involving $a(x)$, $d(x)$, and $c(x)$ comes in --- your $a(x)$ is my $f(x)$.)

Now the containment of ideals $(p(x))\subseteq (f(x))$ implies that $f(x)$ divides $p(x)$. But $p(x)$ is irreducible --- so this is only possible if $f(x)$ is a scalar multiple of $p(x)$, or if $f(x)$ is constant. These cases correspondingly imply $(f(x))=(p(x))$, or $(f(x))=F[x]$.

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