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Let $R$ be a Euclidean domain. An element $u \in R$ is said to be a unit if there is a $v \in R$ such that $uv = 1$. The set of units in $R$ is denoted by $U(R)$.

Let $R$ be a commutative unital ring. We say that an element $u \in R$ is a unit if $u$ has a multiplicative inverse. Prove that the set $U (R) = \{u \ R \colon \; $u$ \text{ is a unit} \}$ is a group under multiplication.

Certainly $1$ is a unit in $R$, so $U(R)$ contains an identity element. Associativity and commutativity of multiplication is inherited from $R$. Let $u \ U(R)$. Then there is $u′ \in R$ such that $uu′ = 1$. Hence $u′$ is a unit, so $u′ \in U(R)$ and $U(R)$ is closed under inverses. Now let $u,v \in U(R)$. Therefore there are $u′,v′ \in R$ such that $uu′ = 1$ and $vv′ = 1$. Since $uvu′v′ = 1$, it follows that uv is a unit in $R$, so $U(R)$ is closed under multiplication. Hence $U(R)$ is a group.

After proving all of this then my question is: Let $F$ be a field. What are the elements of $U(F[x])$ ?

quid
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BeccaBee
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2 Answers2

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The questions asks you for which polynomial $p(x)$ you can find another polynomial $q(x)$ such that their product is $1$.

Certainly neither $p$ nor $q$ can be $0$.

Now, consider what $pq= 1$ says about the degree of the polynomial $p$.

A moments thought will reveal that $p$ has to have degree $0$, that it is a constant polynomial.

So, $U(F[X])$ is a subset of the non-zero constant polynomials, that is, $F \setminus \{0\}$.

It remains to consider which elements of $F \setminus \{0\}$ indeed has a multiplicative inverse. Since $F$ is a field it follows readily that each element of $F \setminus \{0\}$ has a multiplicative inverse.

Thus, $U(F[X]) = F \setminus \{0\}$.

This argument readily generalizes to show that for $R$ an integral domain $U(R[X]) = U(R)$. Note that one uses that $R$ is an integral domain when concluding from $pq=1$ that $p$, $q$ are constant.

For a more general result see Characterizing units in polynomial rings

quid
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We claim that $$U(F[x])=U(F)=F-\{0\}.$$ Of course, $U(F) \subseteq U(F[x])$. Conversely, if $p(x) \in U(F[x])$, then by definition there exists $q(x) \in F[x]$ such that $$p(x)q(x)=1.$$ In particular: $$\deg(p(x)q(x))=\deg(1)$$ $$\deg(p(x))+\deg(q(x))=0,$$ which forces $\deg(p(x))=\deg(q(x))=0$. Hence, $p(x)$ and $q(x)$ are constant polynomials; i.e, $p=p(x),q=q(x) \in F \subset F[x]$. Since $pq=1$, it follows that $p \in U(F)$.

Remark. Observe that the same argument shows that if $R$ is an integral domain, then $$U(R[x])=U(R).$$

Don
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