Groebner basis and prime ideals.

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.

• I'm guessing that's supposed to be $I \cap K[x] \neq (0)$. Because if $I \cap K \neq 0$, then $I$ contains a nonzero element of $K$, which is a unit, so $I = K[x,y_1, \ldots, y_n]$. – André 3000 Feb 5 '16 at 5:00
• That is a typo, thank you. I missed. – Hawk Feb 5 '16 at 5:05
• $G\cap K[X]$ is a groebner basis for the intersection ideal, by the theory on elimination. How are Groebner bases for ideeals in the PID $k[x]$? Also, if $I$ is prime so is its intersection with $k[x]$, as you can easily check. – Mariano Suárez-Álvarez Feb 5 '16 at 5:09
• I did, I can bring it back. But it looks like it been resolved on the end (this might take a while I check my history). – Hawk Feb 12 '16 at 8:53
• @MarianoSuárez-Alvarez, for the second part, is it because $I \cap k[x]$ is prime, so it is maximal in the poly nominal ring an therefore it is irreducible.? – Hawk Feb 12 '16 at 11:28