# Prove that $R[x]$ is an integral domain if and only if $R$ is an integral domain. [duplicate]

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 $$g(x)=0$$.

• Hints: any subring of a domain is a domain, and $R$ is isomorphic to the subring of $R[X]$ of constant polynomials. Jan 13, 2018 at 23:49
• 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 $R[x]$.
– Javi
Jan 13, 2018 at 23:49

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.