Groebner basis and prime ideals.

Question

Let $I$ be an ideal in a polynomial ring $P = K[x,y_1,\dots,y_n]$ and assume that $I \cap K[x]\neq (0)$. Let $>$ be an elimination ordering for $\{y_1, \dots, y_n\}$ and $G$ is a Groebner basis for $I$ with respect to $>$.

1. Why $G\cap K[x]$ is in fact a single polynomial f?

edit: yes I agree $G \cap K$ is also a Grobner 2. And why if $I$ is prime ideal, then $f$ is irreducible?

edit2: $I \cap k[x]$ is a nontrivial prime ideal, so it is maximal, and therefore $f$ is irreducible.