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I'm working on Fulton Algebraic Curves exercise 2-14. I'm stuck on something that I think should be simple but I'm not seeing. The problem is as follows: A set $V \subseteq \mathbb{A}^n$ is called a linear subvariety if $V = \mathcal{V}(f_1,\dots,f_r)$ for $f_i$ of degree $1$. Show that if $T$ is an affine change of coordinates, then $V^T:= T^{-1}(V)$ is also a linear subvariety.

From the definitions we have that $$V^T = T^{-1}(V) = \mathcal{V}(\mathcal{I}(V)^T) = \mathcal{V}(\mathcal{I}(\mathcal{V}(f_1,\dots,f_r))^T) = \mathcal{V}(\mathrm{rad}(f_1,\dots,f_r))^T)$$

I would like to say that this is then equal to $\mathcal{V}((f_1,\dots,f_r)^T) = \mathcal{V}(f_1\circ T,\dots, f_r\circ T)$ which is then evidently a linear subvariety since $T$ and each $f_i$ are degree $1$ polynomials, so the compositions $f_i\circ T$ are also degree $1$. But I don't see why the ideal $(f_1,\dots, f_r)$ is necessarily radical.

Any help would be appreciated.

Noah Solomon
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1 Answers1

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If $f_1, ..., f_n$ are linear, then $I = (f_1, ..., f_n)$ must be prime or all of $k[x_1, ..., x_n]$. In either case $I$ is radical.

For a proof idea of this fact, see: Ideal generated by linear polynomials is radical?

Or, there are other proof strategies and a complete answer to 2-14 here: Ideals in polynomial rings generated by linear polynomials

Lizzie Pratt
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