the ideal of a variety

guachu · **Joined:** Mon, 14 May 2012 20:59:26 UTC **Posts:** 35

Can someone explain me how can i find I(V(I)) if I is an ideal?
I know the definition of the ideal of a variety:
For a variety V is $I(V)=\{f \in K[x], f(p)=0, \forall p \in V\}$

but when I find the points of th set V(I), then, i don't know how find I(V(I)) :confused:

outermeasure · **Posted:** Tue, 15 May 2012 16:48:04 UTC

guachu wrote:

Can someone explain me how can i find I(V(I)) if I is an ideal?
I know the definition of the ideal of a variety:
For a variety V is $I(V)=\{f \in K[x], f(p)=0, \forall p \in V\}$

but when I find the points of th set V(I), then, i don't know how find I(V(I)) :confused:

Recall Hilbert's Nullstellensatz.

guachu · **Joined:** Mon, 14 May 2012 20:59:26 UTC **Posts:** 35

But, for example, you have the ideal I=(x^3+x-y)

and you want to find V(I) and I(V(I)).
V(I) is easy i think, because you have to look for the points (x,y)

so that

that are the points (x,x^3+x)

. But what would you do to calculate I(V(I))?

Shadow · **Posted:** Tue, 15 May 2012 21:02:09 UTC

guachu wrote:

But, for example, you have the ideal I=(x^3+x-y)

and you want to find V(I) and I(V(I)).
V(I) is easy i think, because you have to look for the points (x,y)

so that

that are the points (x,x^3+x)

. But what would you do to calculate I(V(I))?

Well certainly $I(V(I))\subseteq I$ , indeed the Nulstellensatz tells you that $I(V(I))=\sqrt{I}$ . Now you just need to compute $\sqrt{I}$ which is...

guachu · **Joined:** Mon, 14 May 2012 20:59:26 UTC **Posts:** 35

I thought it necessary to work in an algebraically closed field to apply Nulstellensatz, and if the ideal is in R[x,y], R isn't algebraically closed field...
And... why you know that $I(V(I)) \subset I$ ?? I know that $\subset I(V(I))$ is true, but the reverse... why?

Shadow · **Posted:** Wed, 16 May 2012 01:07:29 UTC

guachu wrote:

I thought it necessary to work in an algebraically closed field to apply Nulstellensatz, and if the ideal is in R[x,y], R isn't algebraically closed field...
And... why you know that $I(V(I)) \subset I$ ?? I know that $\subset I(V(I))$ is true, but the reverse... why?

Oops, it seems I accidently wrote those backwards. Aplogies, you are correct.

outermeasure · **Posted:** Wed, 16 May 2012 05:41:40 UTC

guachu wrote:

I thought it necessary to work in an algebraically closed field to apply Nulstellensatz, and if the ideal is in R[x,y], R isn't algebraically closed field...
And... why you know that $I(V(I)) \subset I$ ?? I know that $\subset I(V(I))$ is true, but the reverse... why?

Then look at the version of Nullstellensatz over Jacobson rings. Of course, you should know about schemes before you try out the algebraic-geometry equivalent.

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the ideal of a variety

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