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Show that if $R$ is an integral domain and $f(x)$ is a unit in the polynomial ring $R[x]$, then $f(x) \in R$.

Proceed by contraposition. Suppose $R$ is an integral domain, $f(x)$ is a unit in $R[x]$, but $f(x) \not\in R$. We show $f(x)$ is not a unit in $R[x]$.

Toward a contradiction, suppose $f(x)$ is a unit in $R[x]$, then there exists $f(x)^{-1} \in R[x]$ such that

$$f(x)f(x)^{-1}=1$$

Now, by the division algorithm,

$$\deg(f(x)f(x)^{-1}) = \deg(f(x)) + \deg(f(x)^{-1}) = \deg(1)=0$$

$$\implies\deg(f(x)) = \deg(f(x)^{-1})=0$$

Hence, $f(x) =a$ for some $a \in R$ but contradicts our assumption that $f(x) \not\in R$. Thus, $f(x)$ is not a unit in $R[x]$. By contraposition, the proof is complete.

Does this look good? I don't think there are any jumps in conclusions here but I want to make sure the order of logic is correct here; this proof combines contraposition and contradiction. Also, the fact that $R$ is an integral domain was not used.

EDIT: Based on the comments, a better (and direct) proof would go as follows:

$R$ is an integral domain $\implies R[x]$ is an integral domain$^{(\star)}$. If $f(x) \in R[x]$ is a unit of degree $n$ with inverse $f(x)^{-1}$ of degree $m$, then by $(\star)$,

$$\deg(f(x)f(x)^{-1})= \deg(f(x)) + \deg(f(x)^{-1}) = n+m = \deg(1)=0$$

$$\implies \deg(n)=\deg(m)=0 \implies f(x) \in R$$

$$\hspace{10cm} \blacksquare$$

Aram Nazaryan
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  • This proof is overkill. You can just use the fact that R has no zero divisors to show that a degree n polynomial times a degree m polynomial is a degree n+m polynomial (as the leading coefficients multiply to a non-zero element). This immediately proves the theorem. – kabel abel May 24 '24 at 16:45
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    You are using the fact that $R$ is an integral domain precisely when you say that the degree of the product is the sum of degrees. This is not true for non-integral domains. Otherwise the proof looks okay (although i do not understand why you invoke division algorithm) – nelynx May 24 '24 at 16:52

1 Answers1

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The proof is fine, but overkill. First I recommend some stylistic adjustments:

  • Writing "Toward a contradiction, suppose $f(x)$ is a unit in $R[x]$" is not required since you have already assumed (in the previous line) that $f(x)$ is a unit in $R[x]$.
  • There is no need for a proof by contradiction in the first place. In your argument, you get $f(x) = a \in R$ in the end; that is all you need to show.

The statement $$\deg (f\cdot g) = \deg f + \deg g, \tag{1}$$ holds if $f,g\in R[x]$ and $R$ is an integral domain. For the record, if $R$ is not an integral domain, then $(1)$ may fail, but we have $$\deg(f+g)\leq \max(\deg f, \deg g), \tag{2}$$ and $$\deg (f\cdot g)\leq \deg f +\deg g. \tag{3}$$

stoic-santiago
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