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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 $I=(X,Y)\cap(X-1,Y)\cap (X,Y-1)$.

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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.]

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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 $K[X_1,\ldots,X_n]=e_{i_0}K[X_1,\ldots,X_n]$.

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