Proving this fact about algebraic sets.

Asked Feb 06 '16 at 00:07

Active Feb 06 '16 at 01:08

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I want to prove the following equivalence:

Let $V$ an algebraic set, $K$ a field and $\overline K$ its algebraic closure. Then we say that $V/K$ ($V$ is defined over $K$) if $I_{V}$ (the ideal attached to $V$) is of the form $<S>$ where $S \subset K[x_1,...,x_n]$.

I want to prove that $V/K \Leftrightarrow I_V=I_{V/K}\overline {K}[x_1,...,x_n]$

but the thing is that I have to prove that $I_{V/K} \overline {K}[x_1,...,x_n] \subset K[x_1,...,x_n]$ and that is not true.

On the other hand I have to assume that $I_V=<S>$ but then conclude that $I_V=I_{V/K}\overline {K}[x_1,...,x_n]$ but the thing is that $S \subset K$ so how can I get something depending on $\overline {K}$ sin we know that $K \subset \overline {K}$

So Can someone help me with this please? or is there something wrong in the problem I want to solve ?

Thanks a lot in advance.

edited Feb 06 '16 at 01:08

user26857

52,094

asked Feb 06 '16 at 00:07

user162343

3,245

This simply means $I(V)$ is an ideal in $\overline K[x_1,\dots,x_n]$ with generators that have coefficients in the subfield $K$. – Bernard Feb 06 '16 at 00:11
Can you clarify please ? – user162343 Feb 06 '16 at 00:13
Then $S= I_{V/K}·\overline{K}[x_1,...,x_n]$ and we are done with the first part right? – user162343 Feb 06 '16 at 00:20
That is correct. – Bernard Feb 06 '16 at 00:23
jajajaj perfect, that was something I was thinking but then I was searching for a verfication, thanks, and for the second part, there is something wrong right? – user162343 Feb 06 '16 at 00:25
I'm not sure I interpret correctly you notations, but I suppose $I_V$ the denotes the set of polynomials in $\overline K[x_1,\dots,x_n]$ which vanish on $V$, and $I_{V/K}$ the analogous set in $K[x_1,\dots,x_n]$? – Bernard Feb 06 '16 at 00:31
Yes, in fact I had other notation but I changed it because I thought this was more standard, If you want I can change it. – user162343 Feb 06 '16 at 00:33
No, no, it's OK. If the generators are in $K[x_1,\dots,x_n]$, then $I_V=I_{V/K}\overline K[x_1,\dots,x_n]$, and conversely, $I_{V/K}=I_V\cap K[x_1,\dots,x_n]$ (by faithful flatness of $\overline K[x_1,\dots,x_n]/K[x_1,\dots,x_n]$). – Bernard Feb 06 '16 at 00:38
Can you elaborate more in this fact please?,because I didn't understand this last comment – user162343 Feb 06 '16 at 00:40
Do you know what a faithfully flat module is? – Bernard Feb 06 '16 at 00:44
No ,sorry, I am in second year . – user162343 Feb 06 '16 at 00:45
What I can see is that it is like a "division", right? – user162343 Feb 06 '16 at 00:46
Let us continue this discussion in chat. – Bernard Feb 06 '16 at 00:47
http://math.stackexchange.com/questions/83731/extension-and-contraction-of-ideals-in-polynomial-rings – user26857 Feb 06 '16 at 01:09

Proving this fact about algebraic sets.

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