Obstruction to tensor product decomposition I'm wondering if anything can be said about the following. Suppose $k \subset K \subset R$ are rings (the case where $k$ and $K$ are fields suffices for me). Is it possible to write $R = K \otimes_k S$ for some $k \subset S \subset R$?
My intuition says that this is captured by something like the non-vanishing of some $\text{Tor}_1$, because 1. decomposition problems like these often have cohomological obstructions, and 2. the situation here seems vaguely "dual" to that of expressing $R$ as an extension of $K$ by $S$, which is an element of $\text{Ext}^1(K,S)$. On the other hand, Tor is about modules, not algebras, and even for modules I'm not sure if this intuition is correct.
Example: $\mathbb F_{q^2} \subset \mathbb F_{q^4}$ is not a tensor factor (over $\mathbb F_q$), but $\mathbb F_{q^2} \subset \mathbb F_{q^6}$ is; can this be explained by the (non-)vanishing of some obstruction?
 A: There are some very general things one can say but to keep things simple let's only consider the case that $k \to K$ is a finite Galois extension with Galois group $G$. In this setting, for algebras (or Lie algebras, etc.) such that $R \cong K \otimes_k S$, $S$ is said to be a $k$-form of $R$, and given that a $k$-form exists, they are classified by Galois cohomology; this is the subject of Galois descent.
The condition for when a $k$-form exists is somewhat different, though. Observe that an algebra of the form $K \otimes_k S$ admits a $k$-linear action of the Galois group $G$ acting on the left, which is of course not $K$-linear; the interaction between the $K$-module structure and the Galois action is instead that if $a \in K, s \in S, g \in G$ then
$$g(a \otimes s) = g(a) \otimes g(s).$$
An action of the Galois group $G$ on a $K$-algebra $R$ satisfying this modified version of $K$-linearity is said to be semilinear. So, a necessary condition for the $k$-form $S$ to exist is that $R$ admits a semilinear action of $G$; if this action comes from a $k$-form $S$ then $S$ itself can be recovered as the subalgebra of fixed points of the action of $G$. The simplest version of Galois descent in this case says that this necessary condition is sufficient; given such an action on $R$, the subalgebra $R^G$ of fixed points is a $k$-form of $R$, and all $k$-forms arise in this way.
So, the obstruction is the existence of a semilinear Galois action. I'm not aware of a way to phrase this in cohomological terms; we in fact need such an action already to define the Galois cohomology set that classifies $k$-forms.
Example. Let $k = \mathbb{R}, K = \mathbb{C}$, so the question is when a $\mathbb{C}$-algebra admits a real form. This question is particularly interesting if instead of considering associative algebras we consider Lie algebras, since a key part of the classification of, say, semisimple Lie groups involves the classification of real forms of semisimple Lie algebras. Since $\text{Gal}(\mathbb{C}/\mathbb{R}) \cong C_2$ is cyclic of order $2$ generated by complex conjugation, the definition of a semilinear Galois action reduces to something fairly straightforward in this case: it is an involution $c : \mathfrak{g} \to \mathfrak{g}$ on a complex Lie algebra $\mathfrak{g}$ which is antilinear in the sense that
$$c(aX) = \overline{a} c(X)$$
where $a \in \mathbb{C}, X \in \mathfrak{g}$.
For example, the smallest complex semisimple Lie algebra $\mathfrak{sl}_2(\mathbb{C})$ has exactly two real forms, namely $\mathfrak{sl}_2(\mathbb{R})$ and $\mathfrak{su}(2)$. The corresponding antilinear involutions are complex conjugation $X \mapsto \overline{X}$ in the first case and, more interestingly, negative conjugate transpose $X \mapsto - \overline{X}^T$ in the second case. These real forms can also be thought of as the orthogonal Lie algebras $\mathfrak{so}(2, 1)$ and $\mathfrak{so}(3)$ respectively, reflecting the exceptional isomorphism $\mathfrak{sl}_2(\mathbb{C}) \cong \mathfrak{so}_3(\mathbb{C})$.
