$V(f)\subset V(f,g)$?

Question

I didn't understand this part in this proof:

For me, we have to have the contrary $V(f,g)\subset V(g)$. Maybe the author made a mistake.

Thanks in advance.

@JesseMadnick it's not a book, it's some notes of algebraic geometry, it's really good notes for a beginner. — user42912, Aug 14 '13 at 21:25
Yes, they look very good. Can you provide a link or a reference, please? — Jesse Madnick, Aug 15 '13 at 06:47
@JesseMadnick of course: https://www.andrew.cmu.edu/user/calmost/pdfs/pm464_lec.pdf — user42912, Aug 17 '13 at 19:25

score 5 · Accepted Answer · answered Aug 14 '13 at 20:58

5

Generally, yes, you have $V(f,g) \subset V(f)$ and not $V(f) \subset V(f,g)$. But here, there is a special situation:

Let $g \in I(V(f))$.

That means that $g$ vanishes on $V(f)$, or $V(f) \subset V(g)$, and hence here you have $V(f,g) = V(f)$.

answered Aug 14 '13 at 20:58

Daniel Fischer

just another question, why in the corollary $V(f)$ is irreducible? the author didn't mention this in the proof. – user42912 Aug 14 '13 at 21:16
1

I must admit that my knowledge of algebraic geometry is sparse, but iirc, the vanishing set of an irreducible polynomial is irreducible. – Daniel Fischer Aug 14 '13 at 21:19

1 Answers1