Question about affine coordinate changes

Question

Fulton in his book defines affine coordinate changes:

I'm trying to prove the item (b) of this question:

Let's prove using the induction suggestion.

Suppose $V=V(F_1)$, where $F_1=b_1X_1+\ldots+b_nX_n+b_0$, then $V^T=V(F_1^T)=V(F\circ T)$. I've found this change of coordinates:

$$T'= \begin{pmatrix} b_1^{-1} & 0 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & 0 \\ \end{pmatrix} $$ and

$$T''= \begin{pmatrix} -b_1^{-1}b_0 \\ 0 \\ \vdots \\ 0 \\ \end{pmatrix} $$

So, $F_1^T=X_1$ and $V^T=V(X_1)$, the problem is $T'$ I've found is not invertible, is there anything I can do to fix this?

Thanks

Answer 1

If $V$ is linear and defined by a single polynomial equation $a_1x_1+\ldots+a_nx_n=0$, then put $x_1' = a_1x_1+\ldots+a_nx_n$.

Then consider the map $\mathbb A^n \to \mathbb A^n$ given by $x_1 \mapsto x_1'$ and $x_i \mapsto x_i$. This is obviously an isomorphism as long as $a_1 \neq 0$.

In the image of this map, $V$ is defined by $x_1'=0$.

Answer 2

As the OP noticed in his comments, the other answer to this question is incorrect.

Let $F(X) = a_1X_1 + a_2X_2+\dots+a_nX_n + b$. WLOG, suppose that $a_1\neq 0$. Then define $T = (T_1,\dots,T_n)$ by \begin{align*} T_1(X) &= \frac{1}{a_1}X_1 -\frac{a_2}{a_1}X_2-\cdots-\frac{a_n}{a_1}X_n - \frac{b}{a_1}, \\\\ T_i(X) &= X_i, \text{ if }i\geq 2. \end{align*}

Then you can check that $ (F\circ T)(X) = X_1, $ thus $V^T = V(X_1)$.