Maximal ideal $M$ of $R[x]$ gives prime $M\cap R$ of $R$

Question

Intuition tells me that this is true but I am not sure how to prove this.

Attempt: If $M$ is a maximal ideal then $R[X]/M$ is a field. Suppose $M\cap R$ was not prime. Then $R/M\cap R$ is not an integral domain. So there is some $a,b$ non zero such that $ab=0$. But now we can consider these $a,b$ in $R[X]/M$. This contradicts $R[X]/M$ being a field.

Answer 1

A more general result:

Let $f:A\longrightarrow B$ be a homomorphism of commutative rings, and $\mathfrak q$ be a prime ideal in $B$. Then $\mathfrak p=f^{-1}(\mathfrak q)$ is a prime ideal in $A$

Indeed $f$ induces an injective homomorphism $$\bar f:A/f^{-1}(\mathfrak q)\longrightarrow B/\mathfrak q.$$ Now, if $\mathfrak q$ is prime, the quotient $B/\mathfrak q$ is an integral domain, and $A/f^{-1}(\mathfrak q)$ is isomorphic to a subring, which, necessarily, is also an integral domain.