I have the following problem:
Let $R$ be a PID. Let us denote for $P=a_0+a_1 X+\dots+a_n X^n\in R[X]$ that $c(P)=\gcd(a_0,...,a_n)$. Show that for $P\in R[X]$ nonzero $c(P)=1$ iff for every maximal ideal $\mathfrak{p}\subset R$ the morphism $$R[X]\rightarrow R/\mathfrak{p}[X]$$ sends $P$ to a nonzero element.
I first wanted to show $\Leftarrow$ direction, and my Idea was to do it by contraposition. So assume that $c(P)\neq 1$, i.e. $c(P)\not \in R^\times$ so not invertible. Then we know that $(c(P))\subset R$. Since this is a strict ideal we know from the lecture that there is a maximal ideal $\mathfrak{p}\subset R$ such that $(c(P))\subset \mathfrak{p}$. Then I somehow don't see how to conclude, so I think I should show that the image of $P$ is in the same equivalence class as $\mathfrak{p}$ because then it would map to zero and we would get the negation of the second statement.
Remark. In the solution they wrote something like $p\in \mathfrak{p}R[X]$ and that's the negation of the second statement, but I first don't understand how to get there and then also what does $\mathfrak{p}R[X]$ mean.
Can someone help me? It would be nice if we would continue with the Proof above because till this point I understand it.
You are done. Because $(c(P))\subset\mathfrak p$ implies $a_i\in\mathfrak p$ for all $i$, hence $P\text{ mod }\mathfrak p = 0$.