[proofplan] The proof is a direct unpacking of the definition of the trace pairing. We write both values of $B_{K/k}$ as field traces of products in $K$. Since $K$ is a field, multiplication in $K$ is commutative, so the two products have the same trace. [/proofplan] [step:Expand the trace pairing on both ordered pairs] Let $\alpha,\beta \in K$. By the definition of the trace pairing $B_{K/k}:K \times K \to k$, we have \begin{align*} B_{K/k}(\alpha,\beta)=\operatorname{Tr}_{K/k}(\alpha\beta). \end{align*} Similarly, \begin{align*} B_{K/k}(\beta,\alpha)=\operatorname{Tr}_{K/k}(\beta\alpha). \end{align*} [guided] Fix arbitrary elements $\alpha,\beta \in K$. The trace pairing is the map \begin{align*} B_{K/k}: K \times K \to k \end{align*} defined by sending an ordered pair $(\gamma,\delta) \in K \times K$ to the field trace of the product $\gamma\delta$: \begin{align*} B_{K/k}(\gamma,\delta)=\operatorname{Tr}_{K/k}(\gamma\delta). \end{align*} Applying this definition first to the ordered pair $(\alpha,\beta)$ gives \begin{align*} B_{K/k}(\alpha,\beta)=\operatorname{Tr}_{K/k}(\alpha\beta). \end{align*} Applying the same definition to the reversed ordered pair $(\beta,\alpha)$ gives \begin{align*} B_{K/k}(\beta,\alpha)=\operatorname{Tr}_{K/k}(\beta\alpha). \end{align*} Thus the desired symmetry reduces to comparing the two products $\alpha\beta$ and $\beta\alpha$ inside the field $K$. [/guided] [/step] [step:Use commutativity of multiplication in $K$] Since $K$ is a field, multiplication in $K$ is commutative. Therefore \begin{align*} \alpha\beta=\beta\alpha. \end{align*} Applying the function $\operatorname{Tr}_{K/k}:K \to k$ to this equality gives \begin{align*} \operatorname{Tr}_{K/k}(\alpha\beta)=\operatorname{Tr}_{K/k}(\beta\alpha). \end{align*} Combining this equality with the two expansions above yields \begin{align*} B_{K/k}(\alpha,\beta)=B_{K/k}(\beta,\alpha). \end{align*} Because $\alpha,\beta \in K$ were arbitrary, the trace pairing is symmetric on $K \times K$. [/step]