I want to prove that for a ring $R$, $R[x]$ is an integral domain if and only if $R$ is an integral domain.

I have one direction of the proof ($R$ an integral domain implies $R[x]$) an integral domain, but I am having trouble proving the other direction.

Here is the first direction:

Let $R$ be an integral domain, and suppose on the contrary that $R[x]$ is not an integral domain, meaning that it has zero divisors. In particular, let $f(x),g(x)\neq 0\in R[x]$ such that $\deg(f)=n$ and $\deg(g)=m$ and $f(x)g(x)=0$. Specifically, $f(x)=a_nx^n+...a_1x+a_0$ and $g(x)=b_mx^m+...b_1x+b_0$ where $a_i,b_j\in R$ for all $0\leq i,j\leq n,m$. Now, consider the polynomial term $a_nb_mx^{n+m}\in f(x)g(x)$. Because $f(x)g(x)=0$, this means that each polynomial coefficient must be zero, so in particular $a_nb_m=0$. But, $R$ is an integral domain so either $a_n=0$ or $b_m=0$, contradicting the degree of $f(x)$ or $g(x)$. Thus, $R[x]$ is an integral domain.

And here is what I have so far for the other direction, but I'm not sure if I have the right set-up. Any hints or comments are appreciated.

Now, suppose conversely that $R[x]$ is an integral domain, and suppose on the contrary that $R$ is not, meaning that it has zero divisors. In particular, let $ab=0$ where $a\neq 0$ and $b\neq 0$ for $a_0,b_0\in R$. Furthermore, consider the polynomial product $f(x)g(x)=0$ for $f(x),g(x)\in R[x]$, where $f(x)$ and $g(x)$ are defined in the same way as above. Because $R[x]$ is an integral domain, either $f(x)=0$ or $g(x)=0$.

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    $\begingroup$ Hints: any subring of a domain is a domain, and $R$ is isomorphic to the subring of $R[X]$ of constant polynomials. $\endgroup$ Jan 13, 2018 at 23:49
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    $\begingroup$ I don't see why you're invoking those polynomials at the end. If $R$ has zero divisors, then they are also zero divisors as constant polynomials in $R[x]$. $\endgroup$
    – Javi
    Jan 13, 2018 at 23:49

1 Answer 1


The first direction of the proof looks good - it gets the right idea that the leading coefficient of the product is the product of the leading coefficients of the factors - which of course is non-zero so long as neither factor was zero.

For the second direction, you just need to note that $R$ embeds into $R[x]$ as the constant polynomials. That is, for any $a,b\in R$, you can consider the polynomials $f(x)=a$ and $g(x)=b$. Since $R[x]$ is an integral domain, $f(x)g(x)\neq 0$, but of course, the left hand side is just $ab$ as a polynomial - implying that $ab$ is not zero in $R$ either. More generally, any subring of an integral domain is an integral domain by this reasoning.


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