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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!!!

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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 $R$).

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  • $\begingroup$ Does the unit has inverse? $\endgroup$
    – Minh
    Mar 12, 2019 at 8:25
  • $\begingroup$ 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. $\endgroup$
    – Wuestenfux
    Mar 12, 2019 at 8:30
  • $\begingroup$ In my opinion, an element $f \in R$ is a unit if $\forall g \in R$, we have $fg = 1$. $\endgroup$
    – Minh
    Mar 12, 2019 at 8:32
  • $\begingroup$ 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. $\endgroup$
    – Minh
    Mar 12, 2019 at 8:38
  • $\begingroup$ @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. $\endgroup$
    – Arthur
    Mar 12, 2019 at 8:44

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