Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The ideals $I=(X,Y)$ and $J=(X^2+Y^2)$ in $\mathbb R[X,Y]$ are such that $V(I)=V(J)$ and their radicals aren't the same contradicting the Nullstellensatz (in case it was true for arbitrary fields). However, this shouldn't be a surprise, if we look at their varieties in $\mathbb C$ we find that they aren't the same as the second has a pair of lines that were hidden.

My question is if it is true the other way around: Suppose we have two ideals $I$ and $J$ of $\mathbb K[X_1,\ldots,X_n]$ where $\mathbb K$ is an arbitrary field. Such that $V(I)=V(J)$ and such that ${\bf V}(I)={\bf V}(J)=V(I)$; where ${\bf V}$ means the variety in the affine space of dimension $n$ over $ \mathbb{\bar K}$, the algebraic closure of $\mathbb K$. That is, there are no additional points hidden in the algebraic closure. Is it true then that $\sqrt I=\sqrt J$?

My motivation for the question is a problem of primary descomposition, where we had to find one for $J=(X + Y − X^2 + XY − Y^2,X(X + Y − 1))$ in $\mathbb K[X,Y]$. We had already proved that $V(J)=\{(0,0),(1,0),(0,1)\}$ over $\mathbb{ A}_{\mathbb K}^2$ for any field $\mathbb K$. It made some computations easier in the case of algebraically closed fields because we had automatically that $\sqrt J=I$ where $I$ was the ideal of the three points $I=(X,Y)\cap(X-1,Y)\cap (X,Y-1)$.

share|improve this question

2 Answers 2

up vote 8 down vote accepted

There is a form of the Nullstellensatz (see the wikipedia entry) which is valid for arbitrary fields (and even for more general rings --- so-called Jacobson rings). One way to formulate it (in the case of an arbitrary field) is as follows:

If $k$ is a field, then:

  1. For any ideal $I$ in $k[x_1,\ldots,x_n]$, the radical $\sqrt{I}$ is the intersection of all maximal ideals $\mathfrak m$ in $k[x_1,\ldots,x_n]$ containing $I$.

  2. If $\mathfrak m$ is a maximal ideal of $k[x_1,\ldots,x_n]$, then there is a homomorphism of $k$-algebras $k[x_1,\ldots,x_n] \to \overline{k}$ whose kernel is precisely $\mathfrak m$.

These two facts taken together say that you can recover $\sqrt{I}$ by knowing all of the points of what you call ${\mathbf V}(I)$. (With a little work one can deduce these statements from the Nullstellensatz for $\overline{k}$.)

[Note also that many (I would guess most) people, when they write $V(I)$, would mean what you call ${\mathbf V}(I)$; i.e. even when the ideal consists of polynomials in a non-algebraically closed field $k$, they would take the variety $V(I)$ attached to $I$ to consist of all the common zeroes of the polynomials wiht coordinates lying in $\overline{k}$ (not just those with coordinates lying in $k$). Of course this is just a matter of convention; but it is a common convention which it might be helpful to be aware of.]

share|improve this answer

Yes, take $f \in \sqrt{I}$, then $f^n \in I \subset \bar{K}I$ so $f \in \sqrt{\bar{K}I}=\sqrt{\bar{K}J}$ so $f^m \in \bar{K}J$ for some $m$. Now we have to prove that $\bar{K}J \cap K[X_1,\ldots,X_n]=J$, which is obvious once you have taken a basis $(e_i)_i$ of $\bar{K}$ over $K$ with $e_{i_0}=1$: $\bar{K}J = \oplus_i e_i J$ and $K[X_1,\ldots,X_n]=e_{i_0}K[X_1,\ldots,X_n]$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.