Let $L/K$ be a field extension of finite dimension $d$.
Choosing a $K$-basis for $L$ gives rise to a map $\rho\colon L\to M_d(K)$ representing the map $x \mapsto (y\mapsto xy)$.
Is the image $\rho(L)$ cut from $M_d(K)$ by a set of polynomials with coefficients in $K$? If so, is there a nice way to say what these polynomials are?
By trying out examples, I think the answer is Yes. I also feel like I'm missing something simple.
I am only interested in the case where $K$ and $L$ are number fields, but I feel this assumption is probably not needed.
Jyrki is correct that $\rho(L)$ is a vector subspace and so is cut out by linear equations. But we can be more specific, as follows. "Cayley's theorem for rings" asserts that every ring $R$ can be characterized as the ring of endomorphisms of $R$ as a right $R$-module; that is, the set of abelian group homomorphisms $R \to R$ which commute with right multiplication by elements of $R$ is precisely given by left multiplication by elements of $R$.
If $R$ is commutative left and right multiplication agree so we get a funny statement: $R \subset \text{End}(R)$ is its own centralizer. If $R$ is a finite-dimensional algebra over a field $K$, we can pick a set of generators $\{ e_1, \dots e_g \}$ of it and consider the corresponding multiplication operators $\rho(e_1), \dots \rho(e_g) : R \to R$, and we get
$$\rho(R) = \{ T \in \text{End}_K(R) : T \rho(e_i) = \rho(e_i) T \}.$$
These are linear equations in the entries of $T$ regarded as a matrix in $M_d(K)$ as desired. It is sort of self-referential though, and we use more equations than necessary ($d^2 g$ equations rather than the minimal $d^2 - d$). If furthermore we are considering a finite extension of fields in characteristic $0$ then by the primitive element theorem we can take $g = 1$.
A nice example to work through explicitly is $\mathbb{C}/\mathbb{R}$, which exhibits $\mathbb{C}$ as the real subspace of $M_2(\mathbb{R})$ commuting with the matrix $J = \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right]$.