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

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 $$J = A$$.

• $(d) = (a,p) = (1)\,$ or $(p)$ by $(p)$ irred, so $\,a\not\in (p)\Rightarrow\, (a,p)=(1)\,$ so $(p)$ is maximal. – Bill Dubuque May 13 '19 at 15:58
• 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}\ \ \$ – Bill Dubuque May 13 '19 at 16:01
• What is $a(x)$ in your proof? And how are you using it in the proof? – Julian Mejia May 13 '19 at 16:03
• 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. – Julian Mejia May 13 '19 at 16:58
• 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 – Bill Dubuque May 13 '19 at 17:39

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.
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 $$(f(x))=F[x]$$.