A maximal ideal $M$ of $R[x]$ contains an element that is not a zero divisor

Question

I am working on the following problem:

Let $R[x]$ be the polynomial ring over a commutative ring $R$ with unit $1$. Let $M$ be a maximal ideal of $R[x]$. Show that some element of $M$ is not a zero divisor.

I first tried to take advantage of the fact that $M$ is a maximal ideal of $R[x]$. This means that $R[x]/M$ is a field. Thus, we can take a polynomial $f(x) \in R[x]$, and a polynomial $g(x) \in M$, and observe that there exists some $h(x) \in R[x]/M$ such that $[f(x) + g(x)][h(x)] = 1$, where $1$ represents the unit element in $R[x]$. How can I proceed from this equation to determine that $ab = 0$ for nonzero $a \in M$ implies $b$ must be the zero polynomial ? I suppose I'm just not seeing the trick here.

Thanks!

score 3 · Accepted Answer · answered Nov 18 '19 at 05:56

3

By convention a polynomial $a_nx^n+\cdots+a_0=0$ implies that $a_n=\cdots=a_0=0$, so in particular $x$ is not a zero divisor. To prove the assertion that $M$ contains a non-zero-divisor, look at the following two cases:

Case 1: If $x\in M$, then $x$ satisfies the assertion.

Case 2: If $x\notin M$, then there exists $f(x)$ such that $$xf(x)=1+g(x)$$ for some $g(x)\in M$. This means that $1-xf(x)=-g(x)\in M$, but it is easy to check that $1-xf(x)$ is a non-zero-divisor. QED

answered Nov 18 '19 at 05:56

Pythagoras

7,079

Hello ! How does it follow in case 2 that $xf(x) = 1 + g(x)$ ? This is the only step that I can't follow. – michiganbiker898 Nov 18 '19 at 08:10
1

@michiganbiker898 - $M+(x)=(1)$ – Pierre-Yves Gaillard Nov 18 '19 at 11:46
1

@michiganbiker898 This is explained by Pierre-Yves Gaillard. Alternatively note that if $x\notin M$, then $\bar x$ is a unit in the quotient field $R[x]/M,$ i.e. $\exists f(x)$ such that $\bar x\overline{f(x)}=\bar 1$, so $xf(x)-1\in M$. – Pythagoras Nov 18 '19 at 19:39

score 1 · Answer 2 · answered Nov 18 '19 at 04:41

It is well-known (see Zero divisor in $R[x]$) that if $g(x)$ is a zero divisor in $R[x]$, then there is some $a\in R\setminus\{0\}$ such that $ag(x)=0$.

Now suppose all elements of $M$ are zero divisors. There must be some $n$ such that $x^n\not\in M$, so there is $f(x)$ such that $$x^n \cdot f(x) = 1 + g(x)$$ for some $g(x)\in M$. Then we can find $a\neq 0$ such that $ag(x)=0$, so: $$ x^n\cdot af(x) = a.$$

But it is not possible that $x^n$ divides a constant polynomial.

A maximal ideal $M$ of $R[x]$ contains an element that is not a zero divisor

2 Answers2