I am reading Dummit and Foote's proof that if $R$ is a commutative ring, and if $R[x]$ is a PID, then $R$ is a field. At the start of the proof they write
Proof: Assume $R[x]$ is a Principal Ideal Domain. Since $R$ is a subring of $R[x]$ then $R$ must be an integral domain (recall that $R[x]$ has an identity if and only if $R$ does). ...
I understand how $R$ has an identity if and only if $R[x]$ does, but I don't see how this tells us that $R$ is an integral domain.
To try to work it out myself, if $a,b\in R$ and $ab=0$ then we need to relate this to the other assumptions, so perhaps we consider the ideal $(a,b)$ in $R[x]$. Then there is some $c \in R$ such that $(c) = (a,b)$ and therefore $c = am+bn$ for some $m,n\in R$. Then
$$ abn = 0 $$
$$ a(c-am) = 0 $$
$$ ac = a^2m $$
This seems to me a dead-end. We are not guaranteed by anything that I can see, that there is a unitary element or a unit in any of these rings.
Remember that PID means Principal Ideal Domain.
If $A\subset B$ is an inclusion of rings and $B$ is a domain, $A$ must be as well because any relation $$ xy=0 $$ in $A$ holds in $B$ too.