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I find this definition in Silverman's book, The Arithmetic of Elliptic Curves: an algebraic set(in $A^n(\bar{K})$) is called defined over $K$ if its ideal can be generated by polynomials in $K[X]=K[x_1,...x_n]$.

But what if at the beginning we are given polynomials in $K[X]$ and the algebraic set is defined by these polynomials? I mean, can the ideal of this algebraic set be generated by polynomials in $K[X]$?

I already know that this is equal to the following question:

If $I\subset \bar K[X]$ can be generated by elements in $K[X]$, can $r(I)$ be generated by elements in $K[X]$?

I find it not so easy as I thought. Can anyone lead me out? Thanks!

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    $\begingroup$ I guess $r(I)$ means the radical closure of $I$ (so that $r(I)/I$ is the radical of $\bar{K}[X]/I$). $\endgroup$
    – YCor
    Commented May 5, 2016 at 13:45
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    $\begingroup$ For $n=1$, if $K$ is a nonperfect field and $t\in K$, $u=t^{1/p}\notin K$, consider the ideal $I$ of $\bar{K}[x]$ generated by $x^p-t$. Then $r(I)$ is generated by $x-u$ and all the elements of the intersection $r(I)\cap K[x]$ belong to the ideal generated by $(X-u)^2$, thus $r(I)$ is not defined over $K$. $\endgroup$
    – YCor
    Commented May 5, 2016 at 13:59

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Since you can also take the radical inside the ring $K[x_1,\ldots,x_n]$ this boils down to the following question: if we have a radical ideal $I$ in $K[x_1,\ldots x_n]$, will its extension in $\bar{K}[x_1,\ldots x_n]$ still be radical. In characteristic $p$ this is problematic, take for example $n=1$, take $t\in K$ to be an element without a $p$-th root of unity, and take $I$ to be the ideal generated by the polynomial $x^p-t$, which is radical, but its extension in the algebraic closure will not be radical. For characteristic zero the situation is different: if your ideal $I$ is radical, then it will be the intersection of all maximal ideals containing it in $K[x_1,\ldots x_n]$. Now, if $m$ is such a maximal ideal, then the extension of $m$ in $\bar{K}[x_1,\ldots x_n]$ will be an intersection of maximal ideals there: this is because in characteristic zero extensions are always separable. It follows that the extension of $I$ is also the intersection of maximal ideals in $\bar{K}[x_1,\ldots x_n]$, and therefore it is radical as well.

An edit: It is known that a radical ideal is the intersection of all prime ideals containing it (in any commutative ring). In case we are talking about f.g. algebras over a field, it will also be the intersection of all maximal ideals containing it. To prove this, we will work in the quotient algebra $A/I$ (where $A$ is the algebra). If we have a non-nilpotent element $f$, then we know that $A_f$ contains a maximal ideal $m$. The quotient $A_f/m$ will then be a finite extension $L$ of $K$ (by Zariski's lemma). We thus get a homomorphism $\phi:A\rightarrow A_f\rightarrow L$ such that $f\notin Ker(\phi)$. But $Im(\phi)$ must be a field, because all rings between $L$ and $K$ are fields, so $ker(\phi)$ is a maximal ideal which does not contain $f$.

About the intersection of extensions of ideals: again, it is not true in general, but it is true here, since our extension is just scalar extension from $K$ to $\bar{K}$. More precisely: if $m$ is an ideal in $K[x_i]$, then its extension in $\bar{K}[x_i]$ will be $m\otimes_K\bar{K}$. If now $V$ is any vector space over $K$, and $W_i$ are subspaces in $V$, then it is true that in $V\otimes_K\bar{K}$ we have $(\bigcap_i W_i)\otimes_K\bar{K} = \bigcap_i (W_i\otimes_K\bar{K})$

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  • $\begingroup$ Why radical ideal must be the intersection of maximal ideals? Shouldn't that be prime ideas? $\endgroup$
    – user unknown
    Commented May 6, 2016 at 7:41
  • $\begingroup$ Also, I don't think the extension of intersection equals intersection of extensions. $\endgroup$
    – user unknown
    Commented May 6, 2016 at 7:42
  • $\begingroup$ I added an explanation about this to the answer. $\endgroup$
    – Ehud Meir
    Commented May 6, 2016 at 8:26
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    $\begingroup$ @Ge Tangli: "Why radical ideal must be the intersection of maximal ideals ?" -> the magic words are "Jacobson ring". $\endgroup$
    – Todd Leason
    Commented May 6, 2016 at 8:50
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    $\begingroup$ If $m$ is a maximal ideal in $A=K[x_i]$, then $A/m=L$ is a finite extension field of $K$. Now, due to separability, $A/m\otimes_k\bar{K}\cong \bar{K}^l$ where $l=[L:K]$. This then implies, by looking on the ring $A/m\otimes_K\bar{K}$, the the extension of $m$ is the intersection of maximal ideals in $A\otimes_K\bar{K}$. $\endgroup$
    – Ehud Meir
    Commented May 6, 2016 at 13:12

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