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Let $K$ be an algebraically closed field, and $\mathbb A^n$ the affine-$n$ variety over it. Suppose that $k$ is an arbitrary subfield of $K$. There are definitions on page 217 of Humphreys' Linear Algebraic Groups:

A subvariety $X$ of $\mathbb A^n$ is $k$-closed if $X$ is the set of zeros of some collection of polynomials having coefficients in $k$.

and

$X$ is defined over $k$ if $\mathscr I(X)$ (the ideal in $K[X]$ vanishing on $X$) is generated by $k$-polynomials.

Humphreys says that these two notions coincide when $k$ is perfect. But if I let $K = \mathbb C$, the field of complex numbers, and $k =\mathbb R$, the field of real numbers, and set $X = \{i, -i \}$, then $X$ is the zero set of $f(x) = x^2 +1$ whose coefficients are in $k$. So, $X$ is $k$-closed. But $\mathscr I(X)$ generated by $x-i$ and $x+i$. Apparently, this ideal could not be generated by polynomials with coefficients in $\mathbb R$. So, $X$ is not defined over $k$. But $k =\mathbb R$ is perfect.

Where am I wrong?

Thanks to everyone.

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It seems to me that $x-i$ is not in $\mathscr I(X)$, and that the ideal generated by $x-i$ and $x+i$ is $\mathbb C[x]$. –  Pierre-Yves Gaillard Mar 5 '12 at 14:34
1  
$I(X)=(x^2+1)$. $x-i$ is not in $I$, since $-i-i\neq 0$ –  wxu Mar 5 '12 at 15:38
    
$X$ is not a variety over $\mathbb{C}$, because it is reducibel. But more relevant: the defining ideal is indeed $(x-i)(x+i)$. –  Hagen Mar 5 '12 at 16:02

1 Answer 1

up vote 2 down vote accepted

OK. If I understand the definition.

If $k$ is not perfect, for example, $k=\mathbb{F}_2(t)$ and $K$ its algebraic closure. Then $X=\{t^{1/2}\}$ is $k$-closed, it is the zero of $x^2-t$; however, it is not defined over $k$, since $I(X)=(x-t^{1/2})$ and it can't be defined over $k$.

But if $k$ is perfect, two notions coincide. It is obvious that $X$ defined over $k$ must be $k$-closed. Another direction, if $X$ is $k$-closed, we need to show $X$ is defined over $k$. It is enough to show that for a radical ideal $I$ of $k[x_1,\ldots,x_n]$, the extended ideal $IK[x_1,\ldots,x_n]$ is still radical. It is equivalent to say $k[x_1,\ldots,x_n]/I\otimes_k K$ is reduced. This is true, since $k$ is perfect, $\operatorname{Spec}k[x_1,\ldots,x_n]/I$ is geometrically reduced.

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Thank you very much! –  ShinyaSakai Mar 6 '12 at 10:33

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