Show by example that R[x] does not have a field of fractions if R is not an integral domain.
I am kinda stumped on this question. I understand that since R is not an integral domain, then R[x] does not have a fraction field because R is not commutative. But I am not sure I understand this fully enough to answer the question. Any help is appreciated!
If $R$ is not an integral domain, it has zero divisors. But $R\subset R[x]$, so $R[x]$ has zero divisors and is not an integral domain. If $R[x]$ had a field of fractions $F$, then $R[x]\subset F$, implying $F$ has zero divisors and is not an integral domain.
Any example will do. So let $R=\Bbb Z_4$. Then $2\cdot 2\equiv 0$, with $2\not\equiv0$. Thus no field can have $\Bbb Z_4$ as a subset\subring. In particular $\Bbb Z_4[x]$ hasn't a field of fractions.
Any $\Bbb Z_n$ with $n$ composite gives an example.
Note: any field is an integral domain, because a zero divisor cannot be a unit.
