## Maximal and prime ideals of a polynomial ring over a PID (2)

Posted: September 21, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
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See part (1) here! Again, we will assume that $R$ is a PID and $x$ is a varibale over $x.$ In this post, we will take a look at the maximal ideals of $R[x].$ Let $I$ be a maximal ideal of $R[x].$ By Problem 2, if $I \cap R \neq (0),$ then $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is irreducible modulo $p.$ If $I \cap R =(0),$ then $I=\langle f(x) \rangle$ for some irreducible element $f(x) \in R[x].$ Before investigating maximal ideals of $R[x]$ in more details, let’s give an example of a PID $R$ which is not a field but $R[x]$ has a maximal ideal $I$ which is principal. We will see in Problem 3 that this situation may happen only when the number of prime elements of $R$ is finite.

Example 1. Let $F$ be a filed and put $R=F[[t]],$ the formal power series in the variable $t$ over $F.$ Let $x$ be a variable over $R.$ Then $I:=\langle xt - 1 \rangle$ is a maximal ideal of $R[x].$

Proof. See that $R[x]/I \cong F[[t,t^{-1}]]$ and that $F[[t,t^{-1}]]$ is the field of fractions of $R.$ Thus $R[x]/I$ is a field and so $I$ is a maximal ideal of $R[x]. \ \Box$

Problem 3. Prove that if $R$ has infinitely many prime elements, then an ideal $I$ of $R[x]$ is maximal if and only if $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is irreducible modulo $p.$

Solution. We have already proved one direction of the problem in Problem 1. For the other direction, let $I$ be a maximal ideal of $R[x].$ By the first case in the solution of Problem 2 and the second part of Problem 1, we  only need to show that $I \cap R \neq (0).$ So suppose to the contrary that $I \cap R=(0).$ Then, by the second case in the solution of Problem 2, $I=\langle f(x) \rangle$ for some $f(x) \in R[x].$ We also know that $R[x]/I$ is a field because $I$ is a maximal ideal of $R[x].$ Since $R$ has infinitely many prime elements, we can choose a prime $p \in R$ such that $p$ does not divide the leading coefficient of $f(x).$ Now, consider the natural ring homomorphism $\psi : R[x] \to R[x]/I.$ Since $I \cap R=(0),$ $\psi(p) \neq 0$ and so $\psi(p)$ is invertible in $R[x]/I.$ Therefore $pg(x)-1 \in \ker \psi = I$ for some $g(x) \in R[x].$ Hence $pg(x)-1=h(x)f(x)$ for some $h(x) \in R[x].$ If $p \mid h(x),$ then we will have $p \mid 1$ which is non-sense. So $h(x)=pu(x) + v(x)$ for some $u(x),v(x) \in R[x]$ where $p$ does not divide the leading coefficient of $v(x).$ Now $pg(x) - 1 =h(x)f(x)$ gives us $p(g(x)-u(x)f(x)) - 1 =v(x)f(x)$ and so the leading coefficient of $v(x)f(x)$ is divisible by $p.$ Hence the leading coefficient of $f(x)$ must be divisible by $p,$ contradiction! $\Box$

Example 2. The ring of integers $\mathbb{Z}$ is a PID and it has infinitely many prime elements. So, by Problem 3, an ideal $I$ of $\mathbb{Z}[x]$ is maximal if and only if $I=\langle p, f(x) \rangle$ for some prime $p \in \mathbb{Z}$ and some $f(x)$ which is irreducible modulo $p.$ By Problem 2, the prime ideals of $\mathbb{Z}[x]$ are the union of the following sets:
1) all maximal ideals
2) all ideals of the form $\langle p \rangle,$ where $p \in \mathbb{Z}$ is a prime
3) all ideals of the form $\langle f(x) \rangle,$ where $f(x)$ is irreducible in $\mathbb{Z}[x].$