Can a conjugate-symmetric matrix over an arbitrary field be diagonalized? Let $K$ be a field with an involution $*$, meaning $*:K\to K$ is a field homomorphism and $(x^*)^*=x$ for all $x\in K$. Let $M$ be a conjugate-symmetric $n\times n$ matrix with entries in $K$, meaning ${M_{ij}=M_{ji}}^*$ for all $1\leq i,j\leq n$.
Is $M$ diagonalizable? If so, must its eigenvalues be of the form $xx^*$? If not, what if we restrict to $*=\text{id}$ and/or char$(K)≠2$?

As might be clear, I'm interested in whether the standard real/complex spectral theorems in finite dimension generalize to arbitrary fields. Note that the above question reduces to some standard results for $(K,*)=(\mathbb R,\text{id})$ or $(K,*)=(\mathbb C,\overline{\phantom{x}}\,)$, for which the answers are "yes."
 A: No, consider $\Bbb{Q}(\sqrt{2})$ and a matrix $\begin{pmatrix} 0 & \sqrt{2} \\ -\sqrt{2} &0\end{pmatrix}.$ This matrix has characteristic equation $\lambda^2+2=0$ so it has eigenvalues $\pm\sqrt{2}i.$
The diagonalization of any conjugate-symmetric $A\in M_{n\times n}(\Bbb{F})$ would have the roots of its characteristic polynomial as the diagonal entries, and we'd need to make sure those are elements of the field $\Bbb{F}.$
A: Let $(K, *) = (\mathbb C, \text{id})$ and $M=\begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix}$. Then $M$ is not diagonalizable.
(See Why a complex symmetric matrix is not diagonalizible?)
It's pretty harsh for the spectral theorem to hold. For example, you need the field to be almost algebraically closed, as argued by @Chickenmancer. In general, it's hard to generalize results from linear algebra over $\mathbb R$ or $\mathbb C$ to other fields if the result depends on positivity, like the dot product or conjugation with the property $x^*x\ge 0$.
A: For arbitrary fields of positive characteristic the answer is no.
with $\text{char }\mathbb K = p$, let $J$ be the $p\times p$ matrix of all ones.  Then $\text{rank}\big(J\big) =1$ and  $\text{trace}\big(J\big) =0$ so $J$ is non-zero nilpotent-- i.e. it is not diagonalizable.  But $J=J^T$ and the any field homomorphism sends $1$ to $1$ so $J$ is conjugate-symmetric.
