The following is a problem in Fulton's Algebraic Curves. Note $k$ is algebraically closed.

If $V=V(F_1,\ldots,F_r)$ is an algebraic set where $F_i$ are linear polynomials in $k[x_1,\ldots,x_n]$ then there exist linear polynomial maps $\lbrace T_1,\ldots,T_n\rbrace$ , $T=(T_1,\ldots,T_n):\mathbb{A}^n(k) \to \mathbb{A}^n(k)$ such that $T^{-1}(V)=V(X_{m+1},\ldots,X_n)$ and $T$ is bijective.

I tried to do it by induction. Base case is easy but I am not able to move any further. Any hints would be nice.

  • $\begingroup$ Assume $n=1,r=1$ and $F(x_1) = x_1.(x_1 - 1)$. Then $V(F_1)$ has two elements, isn'it ?. But it seems to me that any set of the form $V(X_1)$ has just one element. So how could such a map $T$ do exists ? $\endgroup$ – Holonomia Jun 21 '15 at 15:47
  • $\begingroup$ The $F_i$ are assumed linear. $\endgroup$ – Hoot Jun 21 '15 at 15:48
  • $\begingroup$ Anyway, this should somehow just be linear algebra. You just need to translate from the subspace language to the equation language. Probably the part that requires any sort of induction was done back when you read Hoffman-Kunze or whichever book :) $\endgroup$ – Hoot Jun 21 '15 at 15:49

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