# Let $R$ be a domain. Prove that if a polynomial in $R[x]$ is a unit, then it is a nonzero constant (the converse is true if $R$ is a field)

Problem: Let $$R$$ be a domain. Prove that if a polynomial in $$R[x]$$ is a unit, then it is a nonzero constant (the converse is true if $$R$$ is a field).

My attempt: We proof that by contradiction. Suppose $$u \in R$$ be a unit of $$R$$ but $$u$$ is not a nonzero constant, then we have $$\deg (u) > 0$$.

$$\forall f \in R$$, $$\deg (uf) \leq \deg (u) + \deg (f)$$.

On the other hand, $$u$$ is a unit of $$R$$, so $$\deg (uf) = \deg (f)$$.

Associate with the inequality above, we see that the equality hold if and only if $$\deg (u) = 0$$. So we can conclude that $$u$$ is a nonzero constant.

Is my proof correct??? Thanks all!!!

Well, a proof by contradition is not necessary. Suppose $$f$$ is a unit with inverse $$g$$. Then $$fg=1$$. Using degrees, we obtain $$0 = {\rm deg}(1) = {\rm deg}(fg) = {\rm deg}(f) + {\rm deg}(g).$$ The last equality holds since $$R$$ has no zero divisors. As the degree is a nonnegative function, it follows that both, $$f$$ and $$g$$ have degree zero and so are constants (elements of $$R$$).

• Does the unit has inverse?
– Minh
Mar 12, 2019 at 8:25
• Yes, indeed. An element $f\in R$ is a unit if there is an element $g\in R$ with $fg=1=fg$. The inverse $g$ is uniquely determined. Mar 12, 2019 at 8:30
• In my opinion, an element $f \in R$ is a unit if $\forall g \in R$, we have $fg = 1$.
– Minh
Mar 12, 2019 at 8:32
• Note that, we are considering $R$ as a domain, which is a commutative ring has unit $1 \neq 0$ and has the zero-product property. So, the inverse of unit here, i'm not sure it has.
– Minh
Mar 12, 2019 at 8:38
• @Minh Did you mean $\forall g\in R$ we have $fg = g$? That's the definition of $1$, sure. But (perhaps unfortunately) that's not the behaviour we want to capture with the word "unit". An element of a ring is a unit if it's invertible, not necessarily the multiplicative identity. Mar 12, 2019 at 8:44