$R$ integral domain implies $R[x]$ integral domain.

MY attempt:

Take $f = a_0 + a_1x + \cdots + a_nx^n, g = b_0 + \cdots + b_mx^m \in R[x] $. Suppose $f(x) \neq 0 $ and $a_n \neq 0 $ and say $fg = 0 $. We show $g = 0$. Notice after multiplyinf $f$ and $g$ leading exponent is $m + n$ hence $\deg f + g = m + n $. We have

$$ fg = (a_0 + a_1x +\cdots + a_nx^n)( b_0 + \cdots + b_mx^m ) = 0.$$

IF we look at term $a_nb_m x^{m+n} $ in $fg$, then we must have $a_nb_m = 0 $. But we know $a_n \neq 0$, hence $b_m = 0 $ since $R$ is integral domain. Can repeat this process $m$ times until obtaining $g = 0$ which shows indeed $R[x]$ is integral domain.

Is this correct?

  • 1
    $\begingroup$ Yes, alternatively, you can assume $g\neq 0$ and $f\neq 0$, and just conclude that $fg\neq 0$. A direct proof might be easier to parse. $\endgroup$ – Prahlad Vaidyanathan Nov 24 '13 at 12:03

Your idea is correct, but the way you formulate it is rather clumsy. Assuming $a_n\neq0$ implies $f\neq0$, so saying both is a bit strange. Better is to say: suppose $f$ and $g$ nonzero, then you can take $n=\deg f$ and $m=\deg g$, and this ensures (by the definition of degree) that $a_n\neq0$ and $b_n\neq0$.

I am not sure what you mean with $\deg f+g=m+n$; did you mean to say $\deg f+\deg g=m+n$? While definitely true if one assumes $n=\deg f$ and $m=\deg g$ as I did above, it is not in itself a very helpful observation. What you need is that $x^{m+n}$ is the leading power of $x$ in $fg$, but for that you need to argue about its coefficient first. So instead of what you wrote, say that the coefficient of $x^{m+n}$ is $a_nb_m$, which is nonzero in an integral domain, and this leads immediately to the conclusion $fg\neq0$. End of proof, no need to loop over the terms of$~g$.

If you want some more training in this, try to prove that if $R$ is an integral domain then, the formal power series ring $R[[x]]$ (which is defined just like $R[x]$, but without requiring a bound of the degree of the terms) is an integral domain as well. In this case there is no leading term for a nonzero power series, but you can still mimick the proof but looking for the lowest degree term in a product of nonzero formal power series.

  • $\begingroup$ please nobody ever reads this $\endgroup$ – user111072 Dec 7 '13 at 18:07

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