# $(x)$ prime ideal in $R[x]$ iff $R$ integral domain by contrapositive

I've done this proof a few ways and I like this one but since it wasn't the "official" one, I wanted to ask if anyone sees a reason it's invalid. It just makes more sense to me on a concrete level. Thanks!

Statement: $R$ is a commutative ring with unity. Show that (1) $(x)$ is a prime ideal in $R[x]$ if and only if (2) $R$ is an integral domain.

Proof: By contrapositives. NOT (2) $\implies$ NOT (1). Assume $R$ is not an integral domain; i.e. there exist some zero divisors, nonzero $a, b \in R$ such that $ab = 0$. Then consider a counterexample using $p(x) = x+a$ and $q(x) = x + b$ which are both not in the ideal $(x)$, but $$p(x)q(x) = (x+a)(x+b) = x^2 + (a+b)x + ab = x(x + a+b) \in (x).$$ Thus $(x)$ is not a prime ideal.

NOT (1) $\implies$ NOT (2). Assume $(x)$ is not a prime ideal in $R[x]$; i.e. there exist some $f(x), g(x) \in R[x]$ such that neither are in $(x)$ but $f(x)g(x) \in (x)$. To be specific, $$f(x) = a_n x^n + ... + a_1 x + a_0, \quad a_0 \neq 0,$$ $$g(x) = b_n x^n + ... + b_1 x + b_0, \quad b_0 \neq 0,$$ but the term $a_0b_0$ in $f(x)g(x)$ is $0$. Then since $a_0, b_0 \in R$, they are zero divisors in $R$ and $R$ cannot be an integral domain.

Basically I was looking for a way around using the intermediate $R[x]/(x) \cong R$ theorem because although I know it and can prove it, I'm paranoid I won't have time on my qual, and I don't want to just use it without justification so counterexamples seemed faster. If I've broken logic somewhere please let me know!

• Nice job looks good to me; one small mistake though, when you wrote $g(x)$ you also made it of degree $n$ instead of another degree. – Ovi Aug 2 '18 at 18:36
• @Ovi I think that it is not a mistake because, WLOG, we can add many zero coefficients that we want. – Gödel Aug 2 '18 at 18:41
• @Gödel Oh okay that's right – Ovi Aug 2 '18 at 20:08
• Yes, it looks great, +1. – Prism Aug 3 '18 at 5:49
• Great, thanks so much! – metricforbees Aug 3 '18 at 19:38