I have a problem about a line inside the following proof which is actually from the book Abstract Algebra by Dummit & Foote
Proposition (Eisenstein's Criterion). Let $P$ be a prime ideal in the integral domain $R$ and let $~f(x)=x^n+a_{n-1}x^{n-1}+\dots +a_0$ be a polynomial in $R[x]$ $(n\ge1)$. Suppose $a_{n-1},\dots ,a_0$ are all elements of $P$ and $a_0$ is not an element of $P^2$. Then $f(x)$ is irreducible.
Proof. Suppose $f(x)$ is reducible, say $f(x)=a(x)b(x)$ in $R[x]$, where $a(x),b(x)$ are non-constant polynomials. Reducing this equation modulo $P$ and using the assumptions on the coefficients of $f(x)$ we obtain the equation $x^n=\overline{a(x)b(x)}$ in $(R/P)[x]$, where bar denotes the polynomials with coefficients reduced mod $P$. Since $P$ is prime ideal, $R/P$ is an integral domain, and it follows that both of $\overline{a(x)}$ and $\overline{b(x)}$ have $0$ constant term, i.e, the constant term of both $a(x)$ and $b(x)$ are elements of $P$. But then constant term of $a_0$ of $~f(x)$ as the product of these two would be an element of $P^2$, a contradiction. This completes the proof.
But I cannot understand the line which I have bold in the proof. What I understand is that since $x^n=\overline{a(x)b(x)}$, so comparing the coefficients on both sides we get the product of the constant terms of $~\overline{a(x)}$ and $~\overline{b(x)}$ is zero in $(R/P)[x]$, which is a Integral domain.....then either the constant term of $~\overline{a(x)}=0$ or constant term of $~\overline{b(x)}=0$...But how does both of them is zero?
Please help.