Maximal and prime ideals of a polynomial ring over a PID (1)

Posted: September 21, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: Gauss's lemma, irreducible polynomial modulo p, maximal ideals of R[x], PID, prime ideals of R[x]

We know that if $R$ is a field and if $x$ is a variable over $R,$ then $R[x]$ is a PID and a non-zero ideal $I$ of $R[x]$ is maximal if and only if $I$ is prime if and only if $I$ is generated by an irreducible element of $R[x].$ If $R$ is a PID which is not a field, then $R[x]$ could have prime ideals which are not maximal. For example, in $\mathbb{Z}[x]$ the ideal $I:=\langle 2 \rangle$ is prime but not maximal. In this two-part post, we will find prime and maximal ideals of $R[x]$ when $R$ is a PID.

Notation. Throughout this post, $R$ is a PID and $R[x]$ is the polynomial ring in the variable $x$ over $R.$ Given a prime element $p \in R,$ we will denote by $\phi_p$ the natural ring homomorphism $R[x] \to R[x]/pR[x].$

Definition Let $p$ be a prime element of $R.$ An element $f(x) \in R[x]$ is called irreducible modulo $p$ if $\phi_p(f(x))$ is irreducible in $R[x]/pR[x].$ Let $\gamma : R \to R/pR$ be the natural ring homomorphism. Then, since $R[x]/pR[x] \cong (R/pR)[x],$ an element $f(x)=\sum_{i=0}^n a_ix^i \in R[x]$ is irreducible modulo $p$ if and only if $\sum_{i=0}^n \gamma(a_i)x^i$ is irreducible in $(R/pR)[x].$ Note that $R/pR$ is a field because $R$ is a PID.

Problem 1. Prove that if $p \in R$ is prime and if $f(x) \in R[x]$ is irreducible modulo $p,$ then $I:=\langle p, f(x) \rangle$ is a maximal ideal of $R[x].$ If $f =0,$ then $I$ is a prime but not a maximal ideal of $R[x].$

Solution. Clearly $I/pR[x]=\phi_p(I)=\langle \phi_p(f(x)) \rangle.$ So $\phi_p(I)$ is a maximal ideal of $R[x]/pR[x]$ because $\phi_p(f(x))$ is irreducible in $R[x]/pR[x] \cong (R/pR)[x]$ and $R/pR$ is a field. So $I$ is a maximal ideal of $R[x].$ If $f =0,$ then $I=\langle p \rangle=pR[x]$ and so $R[x]/I \cong (R/pR)[x]$ is a domain which implies that $I$ is prime. Finally, $I= \langle p \rangle$ is not maximal because, for example, $I \subset \langle p,x \rangle \ \Box$

Problem 2. Prove that a non-zero ideal $I$ of $R[x]$ is prime if and only if either $I= \langle f(x) \rangle$ for some irreducible element $f(x) \in R[x]$ or $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is either zero or irreducible modulo $p.$

Solution. If $f(x) \in R[x]$ is irreducible, then $\langle f(x) \rangle$ is a prime ideal of $R[x]$ because $R[x]$ is a UFD. If $f(x)=0$ or $f(x)$ is irreducible modulo a prime $p \in R,$ then $I=\langle p, f(x) \rangle$ is a prime ideal of $R[x]$ by Problem 1.
Conversely, suppose that $I$ is a non-zero prime ideal of $R[x].$ We consider two cases.
Case 1. $I \cap R \neq (0)$ : Let $0 \neq r \in I \cap R.$ Then $r$ is clearly not a unit because then $I$ wouldn’t be a proper ideal of $R[x].$ So, since $r \in I$ and $I$ is a prime ideal of $R[x],$ there exists a prime divisor $p$ of $r$ such that $p \in I.$ So $pR[x] \subseteq I$ and hence $\phi_p(I)=I/pR[x]$ is a prime ideal of $R[x]/pR[x] \cong (R/pR)[x].$ Thus we have two possibilities. The first possibility is that $\phi_p(I)=(0),$ which gives us $I \subseteq \ker \phi_p = pR[x]$ and therefore $I=pR[x]=\langle p \rangle.$ The second possibility is that $\phi_p(I)=\langle \phi_p(f(x)) \rangle= \phi_p(\langle f(x) \rangle)$ for some irreducible element $\phi_p(f(x)) \in R[x]/pR[x],$ which gives us $I=\langle p, f(x) \rangle$ because $\ker \phi_p =pR[x].$
Case 2. $I \cap R = (0)$ : Let $Q$ be the field of fractions of $R$ and put $J:=IQ[x].$ Then $J$ is a non-zero prime ideal of $Q[x]$ because $I$ is a prime ideal of $R[x].$ Note that $J=\{g(x)/r : \ g(x) \in I, \ 0 \neq r \in R \}.$ So, since $Q[x]$ is a PID, $J=q(x)Q[x]$ for some irreducible element $q(x) \in Q[x].$ Obviously, we can write $q(x)=\alpha f(x),$ where $\alpha \in Q$ and $f(x) \in R[x]$ is irreducible and the gcd of the coefficients of $f(x)$ is one. Thus $J = f(x)Q[x]$ and, since $f(x) \in J,$ we have $f(x) = g(x)/r$ for some $0 \neq r \in R$ and $g(x) \in I.$ But then $rf(x)=g(x) \in I$ and so $f(x) \in I$ because $I$ is prime and $I \cap R = (0).$ Hence $\langle f(x) \rangle \subseteq I.$ We will be done if we prove that $I \subseteq \langle f(x) \rangle.$ To prove this, let $h(x) \in I \subseteq J=f(x)Q[x].$ So $h(x)=f(x)q_0(x)$ for some $q_0(x) \in Q[x].$ Therefore, since the gcd of the coefficients of $f(x)$ is one, we must have $q_0(x) \in R[x]$ by Gauss’s lemma. Hence $h(x) \in \langle f(x) \rangle$ and the solution is complete. $\Box$

See the next part here!

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