# Suppose $K/k$ is a finite extension of finite fields. Show that both the norm and trace are injective.

Statement: Suppose $$K/k$$ is a finite extension of finite fields. Show that both the norm and trace are injective.

Proof: For a finite extension of fields $$K/k$$, we associate to each element $$\alpha$$ of $$K$$ the $$k$$-linear transformation $$m_\alpha: K\to K$$ where $$m_\alpha$$ is multiplication by $$\alpha$$:

$$m_\alpha(x)=\alpha x$$ for $$x\in K$$. Each $$m_\alpha$$ is a $$k$$-linear function from $$K$$ to $$K$$. We define $$Tr_{K/k}(\alpha) = Tr(m_\alpha)=Tr(\alpha x)=\alpha Tr(x).$$ Assume $$\alpha, \beta \in K$$ where $$\alpha\neq \beta$$. Then $$Tr_{K/k}(\alpha)=Tr_{K/k}(\beta) \iff \alpha Tr(x) = \beta Tr(x) \iff \alpha =\beta$$ Hence the trace is injective.

We define $$N_{K/k}(\alpha)=N(m_\alpha)=\det(\alpha x) =\alpha^n \det(x)$$ where $$n=[K:k]$$. Similarly, we find that the norm is also injective.

Now this seems too easy. Did I do something wrong?

• The norm is $N(a) =\prod_{j=0}^{n-1} a^{q^j} = a^{ (q^n-1)/(q-1)}$ which is injective iff $q^n-1$ and $(q^{n+1}-1)/(q-1)$ are coprime which never happens for $n \ne 1$. The trace is $Tr(a)=\sum_{j=0}^{n-1} a^{q^j}$ which doesn't have such a closed-form, for $p | n$ its kernel contains $F_q$, otherwise $a \mapsto \frac{Tr(a)}{n}$ is a $F_q$-linear projection $F_{q^n}\to F_q$ so its kernel is $n-1$ dimensional. Commented Aug 26, 2019 at 3:48

The statement isn't true. The norm and trace take values in the ground field, so for example the image of $$N_{\mathbb{F}_{49}/\mathbb{F}_7}$$ must send a set of 49 elements to a set of 7, so it cannot be injective.
As for why your argument is wrong, you are thinking of $$\alpha$$ as a scalar when it is not, so you can't pull it out of the trace or determinant like you would an element of the ground field.