Units in Polynomial Rings

We classify the units in polynomial rings.

Proposition. Let \(R\) be a commutative ring, and let \(f\in R[x]\) be a polynomial. Expand \[f(x)=\sum_{i=0}^da_ix^i.\] Then \(f\) is a unit if and only if \(a_0\) is a unit, and \(a_i\) is nilpotent for all \(i>0\).

It is relatively easy to show that all these polynomials are in fact units; this more or less follows from the following lemma.

Lemma 1. Let \(R\) be a commutative ring. If \(u\in R\) is a unit, and \(n\in R\) is nilpotent, then \(u+n\) is a unit.

Proof of Lemma 1. It is enough to show that \(u^{-1}(u+n)=1+u^{-1}n\) is a unit, which reduces us to the case that \(u=1\). Additionally, by replacing \(n\) with \(-n\) (which is still nilpotent), it is enough to check that \(1-n\) is a unit. We will do this by hand: given that \(n^N=0\), we see \[(1-n)\Bigg(\sum_{i=0}^{N-1}n^i\Bigg)=1-n^N\] simply equals \(1\), so we are done. \(\blacksquare\)

It remains to show that the proposition provides all the units of \(R[x]\). Our argument will use the Axiom of Choice (!), though there are more hands-on arguments which do not. The Axiom of Choice enters via the following lemma.

Lemma 2. Let \(R\) be a commutative ring. Then the nilradical ideal equals the intersection of all prime ideals in \(R\).

Proof of Lemma 2. We show the two inclusions separately.

• Suppose that \(a\) is in the nilradical. Then \(a^n=0\) for some \(n>0\). Thus, for any prime ideal \(\mathfrak{p}\), we see that \(a^n\in\mathfrak{p}\), so \(a\in\mathfrak{p}\) follows because \(\mathfrak{p}\) is prime.

• Suppose that \(a\) is not in the nilradical. Then we want to find a prime ideal not containing \(a\). For this, we let \(S\) be the multiplicative set generated by \(a\), and we would like to find a prime ideal avoiding \(S\). Well, simply let \(\mathfrak{p}\) be a maximal ideal avoiding \(S\), which exists by Zorn’s lemma, which applies because the union of a chain of ideals avoiding \(S\) continues to be an ideal avoiding \(S\).

  It remains to show that \(\mathfrak p\) is actually prime. Well, suppose that \(x\) and \(y\) are not in \(\mathfrak p\), and we will show that \(xy\) is also not in \(\mathfrak p\). Because \(x\) and \(y\) are not in \(\mathfrak p\), the ideals \((x)+\mathfrak p\) and \(y+\mathfrak p\) are both strictly larger than \(\mathfrak p\), so they must contain an element of \(S\). The product of these elements of \(S\) then produces an element of \[(xy)+(x)\mathfrak p+(y)\mathfrak p+\mathfrak p^2\subseteq(xy)+\mathfrak p\] and still in \(S\). It follows that \((xy)+\mathfrak p\) is an ideal strictly larger than \(\mathfrak p\), so \(xy\notin\mathfrak p\). \(\blacksquare\)

Remark. Here is an intuitive way to think about the second step: prime ideals of \(R\) avoiding the multiplicative set \(S\) are in bijection with the prime ideals of the localization \(S^{-1}R\) via the natural map \(\operatorname{Spec}S^{-1}R\to\operatorname{Spec}R.\) Thus, the former set is nonempty because the latter set is nonempty: simply take a maximal ideal of \(S^{-1}R\)!

We now begin proving the proposition more directly. To start off, we handle the case of an integral domain separately

Lemma 3. Let \(R\) be an integral domain. Then \(f\in R[x]\) is a unit if and only if \(f\) is constant and \(f(0)\) is a unit.

Proof of Lemma 3. Certainly, if \(f\) is a constant polynomial \(f=a_0\) with \(f(0)=a_0\) a unit, then \(f\) is a unit whose inverse is just \(a_0^{-1}\).

For the converse direction, we use a degree argument. Suppose that \(f\) is a unit with inverse \(g\); certainly both of these elements should be nonzero (or else \(R[x]=0\) and the statement has little content). Furthermore, let \(a_dx^d\) and \(b_ex^e\) be the leading terms of \(f\) and \(g\), where \(a_d\ne0\) and \(b_e\ne0\). Then \(fg\) has as its highest-possible degree term \(a_db_ex^{d+e}\), and because \(R\) is an integral domain, we see that this term is nonzero!

Thus, because \(fg=1\), we conclude that \(d+e=0\) and \(a_db_e=1\). In particular, \(d=0\), so \(f\) is constant, and \(f(0)=a_d=a_0\) is a unit. \(\blacksquare\)

Proof of Proposition. Let’s start with the backward direction. If \(a_0\) is a unit, and \(a_i\) is nilpotent for \(i>0\), then \(f(x)-f(0)\) is nilpotent while \(f(0)\) is a unit, so \(f\) is a unit by Lemma 1.

We now handle the forwards direction. Suppose that \(f\) is a unit. The main claim is that \(a_i\) is nilpotent for all \(i>0\): this would then imply that \(f(x)-f(0)\) is nilpotent while \(f\) continues to be a unit, thereby verifying further that \(f(0)\) is a unit.

It remains to prove the main claim. For this, we reduce modulo primes. Indeed, note that any reduction \(f\pmod I\in (R/I)[x]\) continues to be a unit for any ideal \(I\subseteq R\). In particular, for any prime \(\mathfrak p\), Lemma 3 tells us that all non-constant coefficients \(f\pmod{\mathfrak p}\) must vanish! In other words, \(a_i\in\mathfrak p\) for all \(i>0\) and all primes \(\mathfrak p\). Thus, Lemma 2 implies that each \(a_i\) is nilpotent! \(\blacksquare\)

Remark. Here is an alternate proof of Lemma 1, in a similar spirit as the above proof. It is enough to check that \(u+n\) is not contained in any maximal ideal, for then the ideal \(u+n\) must be all of \(R\). Well, \(u\) is not contained in any maximal ideal because \((u)=R\), and \(n\) is contained in every maximal (in fact, prime) ideal by Lemma 2, so \(u+n\) cannot be contained in any maximal ideal.