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Theorem: Let $R$ be an integral domain, then ring of polynomials $R[x]$ form an integral domain.

To prove this, i first prove a little lemma (not giving the proof here):

Lemma: Let $R$ be integral domain, then $\forall p,q\in R[x]$ we have that $\deg{(pq)}=\deg{(p)}+\deg{(q)}$.

In order to prove the theorem, we have to show that if we have any two nonzero polynomials $p,q$, then their product is nonzero. Because $p,q$ is nonzero, we have that their degree is greater or equal to 0. By lemma $\deg{(pq)}=\deg{(p)}+\deg{(q)}\geq0+0=0$ this shows that $\deg{(pq)}\geq 0$ so $pq$ is nonzero and we conclude that $R[x]$ forms an integral domain.

Is my proof valid or should I aim for something stronger? All the proofs I am aware of always are long examining the terms of the polynomial $pq$ by definition and so on, so I was just wondering, if this argument is correct. Thanks

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  • $\begingroup$ The lemma requires $R$ to be an integral domain. It can be proved by simply comparing leading terms. They cannot cancel. Consider the zero polynomial separately. $\endgroup$
    – Wuestenfux
    Commented Sep 3, 2018 at 15:13
  • $\begingroup$ Having $deg(pq) \geq 0$ could mean $deg(pq)=0$ and does not imply $pq \neq 0$, as degree of zero polynomial is zero $\endgroup$
    – nikola
    Commented Sep 3, 2018 at 15:13
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    $\begingroup$ @nikola well, usually the degree of zero polynomial is defined as $-\infty$ or $-1$. Yes, i forgot to add in the lemma, that we assume $R$ is an integral domain, editing it. $\endgroup$
    – Michal Dvořák
    Commented Sep 3, 2018 at 15:14
  • $\begingroup$ zero polynomial is just another constant, and degree of any constant is zero, making degree of polynomial lower then zero is pointless $\endgroup$
    – nikola
    Commented Sep 3, 2018 at 15:15
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    $\begingroup$ @nikola, the point of defining $\mathrm{deg}(0) = -\infty$ is so that for integral domains you have the rule $\mathrm{deg}(fg) = \mathrm{deg}(f) + \mathrm{deg}(g)$. $\endgroup$
    – Mees de Vries
    Commented Sep 3, 2018 at 15:17

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Is my proof valid or should I aim for something stronger?

Your proof is perfectly valid.

All the proofs I am aware of always are long examining the terms of the polynomial $pq$ by definition and so on, so I was just wondering, if this argument is correct.

I doubt the proofs are that much longer; all the work that goes into examining coefficients of $pq$ will instead go into proving the lemma.

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  • $\begingroup$ Yeah, that's true, the part with examining the coefficients is in the lemma, thanks for the verification! $\endgroup$
    – Michal Dvořák
    Commented Sep 3, 2018 at 15:47

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