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Let $k$ be an algebraic subfield of $C$.

Let's first define affine algebraic set and variety. If $F$ is a subset of $k[X]=k[x_1,...,x_n]$ then the affine algebraic set is defined as follows

$$\{a\in k^n| f(a)=0\ \text{for every}\ f\in F \}$$

Now we define a variety, Var(I) where $I$ is an ideal of $k[X]$, as an irreducible affine algebraic set.

I want to prove Nullstellensatz , $\text{Id}(V(I))=\sqrt{I}$, using the fact that every variety has a k-generic point, https://mathoverflow.net/questions/89368/intuition-behind-generic-points-in-a-scheme

And the definition of k-generic point is: A point $x\in V$ is $k$-generic if every polynomial with values in $k$ that vanishes on $x$, vanishes on all of $V$.

The side $\sqrt{I}\subset\text{Id}(V(I)) $ is obvious. We need to show that $\text{Id}(V(I)) \subset \sqrt{I}$. Let $a$ be the $k$-generic point of $V(I)$. We need to show that if a polynomial $h \in \text{Id}(V(I))$ then there is an integer $m>0$ such that $h^m\in I$.

I am not sure how to use the $k$-generic point $a$.

S_Alex
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  • What do you mean by "let $k$ be an algebraic subfield of $C$"? Do you mean algebraically closed, algebraic over the prime subfield, or something else? Secondly, what is your definition of a $k$-generic point? Thirdly, what is the definition of "variety" that you use? – KReiser Feb 29 '20 at 22:33
  • By algebraic subfield, I mean algebraically closed. By k-generic point of a variety, I mean if a polynomial $f$ vanishes at that point it vanishes at all points of the variety. By variety, I mean irreducible algebraic closed set. – S_Alex Mar 01 '20 at 07:52
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    Please be more specific about your definition of variety: do you mean $\operatorname{Spec} R$ for $R$ a $k$ -algebra with some extra conditions, a closed subset of $k^n$, or something else? There's a wide variety of possible definitions, and knowing what yours is is necessary to write a good answer. – KReiser Mar 01 '20 at 09:08
  • By algebraic closed set I mean, it is a set of common zeros of a set of polynomials, and a variety is an irreducible algebraic closed set. – S_Alex Mar 01 '20 at 19:07
  • This still is not specific enough: for instance, if we take your definition with the assumption that everything is happening inside $k^n$ as a set, there are no generic points - so for your question to make sense, this cannot be the case. But it's not clear from your definition that you exclude this. Please edit in to your post a full and complete definition of "variety" that you're working with. – KReiser Mar 01 '20 at 19:39
  • It is obvious that every subset of $k^n$ is not an algebraic closed set. a set is an algebraic closed set provided that there is a set of polynomials such that that set is the set of common zeros of the set of polynomials. You can see the definition here: https://en.wikipedia.org/wiki/Algebraic_variety – S_Alex Mar 01 '20 at 20:01
  • I think you may be confused: I did not say anything about "every subset of $k^n$" being "an algebraic closed set". I simply requested for you to provide your full definition in the post, and noted that your current attempt leaves open the possibility that you do not have any generic points to talk about, contradicting your stated goal. This is still true with your citation of wikipedia: the (first) definition listed on that page has everything happening inside $k^n$ as sets. – KReiser Mar 01 '20 at 20:07
  • If you take a look at this link: https://mathoverflow.net/questions/89368/intuition-behind-generic-points-in-a-scheme, It says: Definition: Let $k \subset C$ be a subfield of the complex numbers and $V$ an affine complex variety. A point $x \in V$ is generic if every polynomial with values in $k$ that vanishes on $x$, vanishes on all of $V$. Then, Proposition: If $C/K $ has infinite transcendental degree, then every variety $V$ has a $k$-generic point. The link has proof – S_Alex Mar 01 '20 at 20:23
  • I have a PhD in the subject and I have written a fair bit here on this site (and elsewhere!) about what an algebraic variety is. All I am asking is for you to write down exactly what your definition is. Your most recent comment seems to suggest you're using schemes - is that true? – KReiser Mar 01 '20 at 20:25
  • I don't know about schemes. You can also take a look at Rotman's book "Advanced modern algebra" edition 2 page 379. However Rotman does not distinguish between reducible and irreducible sets. I mean irreducible ones. – S_Alex Mar 01 '20 at 20:34
  • Please write your definition in the post. Not everyone has access to this book, and writing your definition in your post will help people to answer your question. Until you do this, I will not be responding further. – KReiser Mar 01 '20 at 20:36
  • I edited the post – S_Alex Mar 01 '20 at 21:04

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