Silverman AEC-Theorem 9.3

Question

Can someone explain why in the last line(image 2),$T\alpha=0\implies T\alpha\beta=0$.

Answer 1

This line is somewhat confusing - recall that Silverman is replacing $\beta$ with $\beta - \frac{1}{2}T(\beta) - \frac{T(\alpha \beta)}{2\alpha^2}\alpha$ in the previous line. So in the line you are asking about, this is the quantity that $\beta$ represents. For clarity I will leave $\beta$ fixed, and set $\beta' := \beta - \frac{1}{2}T\beta - \frac{T(\alpha\beta)}{2\alpha^2}\alpha$. Then the question is, why is $T(\beta') = T(\alpha \beta') = 0$?

The answer is a computation: \begin{align*} T(\beta') & = T(\beta) - \frac{1}{2}T\beta - \frac{T(\alpha \beta)}{2\alpha^2}\alpha \\ & = T(\beta) - \frac{1}{2}T(T(\beta)) - \frac{T(\alpha \beta)}{2\alpha^2}T(\alpha) \\ & = T(\beta) - \frac{1}{2}(2T\beta) - 0 \\ & = 0. \end{align*}

Where we have used the following facts of our situation: the trace is $\mathbb{Q}$-linear, $T(\alpha \beta), T(\beta), \alpha^2 \in \mathbb{Q}$, and $T(q) = 2q$ for any $q \in \mathbb{Q}$, and $T(\alpha) = 0$. The fact that $T(\alpha \beta') = 0$ is a similar computation.