Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $k$.


$X$ 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.

share|cite|improve this question
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
$I(X)=(x^2+1)$. $x-i$ 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 Knaf Mar 5 '12 at 16:02
Some authors define an affine variety to be the ringed space of a closed set, rather than an irreducible closed set. – D_S Oct 10 '15 at 20:22
up vote 3 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.

share|cite|improve this answer
Thank you very much! – ShinyaSakai Mar 6 '12 at 10:33

The ideal corresponding to $\{i,-i\}$ is $$I( \{i\} \cup \{-i\} ) = I \{i\} \cap I \{-i\} = (X-i) \cap (X+i) = (X - i)(X+i) = (X^2 +1)$$ which can be generated by a polynomial in $\mathbb{R}[X]$.

share|cite|improve this answer

Your Answer


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.