0
$\begingroup$

Let $X$ be an affine variety, $X \subset A^n$ and suppose $f_1(T),\ldots,f_r(T) \in K[T_1,\ldots,T_n] $ generate $I(X)$. (Note that $I(X)$ is the ideal of $K[T_1,\ldots,T_n]$ of which elements of $X$ are the common zeroes).

$\operatorname{Tan}(X)_x$ for some $x \in X$ is defined as the common zeroes of $d_xf_j$'s where $f_j$ range over $I(X)$. How do we show that it is equivalent to say that $Tan(X)_x$ is the set of common zeroes of $d_xf_1,\ldots,d_xf_r$ ?

Notation: For $f(T_1,\ldots,T_n)$, we define $d_xf = \sum_{i=1}^n \frac{\partial f}{\partial T_i} (x) (T_i - x_i) $, where $x = (x_1,\ldots,x_n)$ is an element of $X$.

This is an exercise in Humphreys book on linear algebraic groups too (page no- 44, Exercise 1). I was trying to read the theory and was stuck in showing this. Any help is appreciated.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

The following identities can be routinely verified. $$d_x(f+g) = d_x f + d_x g$$ $$d_x (fg) = f(x) d_x g + g(x) d_x f$$ for all $f,g \in k[T_1, \dots, T_n]$

Then the $d_x g_i$, where $g_i \in I(X)$, lie in the ideal $(d_x f_1, \dots, d_x f_n)$. The reverse inclusion is trivial, hence $(d_x g_i) = (d_x f_1, \dots, d_x f_n)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .