TITLE: Real forms of complex vector spaces and $\mathbb{C}$-algebra
QUESTION [11 upvotes]: A real form $W$ of a complex vector space $V$ is a real subspace s.t. $\mathbb{C}\otimes_{\mathbb{R}}W \cong V$ by $a\otimes x \longrightarrow ax$, or equivalently there is an $\mathbb{R}$-basis of $W$ that is also a $\mathbb{C}$-basis of $V$.
There is a fact that every complex vector space has a real form.
For example, $\mathbb{R}^n$, $\mathbb{R}[x]$ and $M_n(\mathbb{R})$ are real forms of $\mathbb{C}^n$, $\mathbb{C}[x]$ and $M_n(\mathbb{C})$ respectively. The last two examples are not only real vector spaces but real subalgebras.
So, I'm wondering whether any complex associative algebra $A$ has a real form, i.e. a real subalgebra $B$ s.t. $\mathbb{C}\otimes_{\mathbb{R}}B \cong A$ as $\mathbb{C}$-algbras? Is there any counterexample?

REPLY [2 votes]: Here is a sketch of a construction:
Let $k$ be a perfect field. The $j$-invariant provides a map
$$j : \{\text{elliptic curves over } k\}/~{\cong} \to k$$
which is bijective when $k$ is algebraically closed. It commutes with base change (since the $j$-invariant can be computed from the Weierstrass equation). Now take an elliptic curve over $\mathbb{C}$ with $j$-invariant in $\mathbb{C} \setminus \mathbb{R}$. Its function field is a $\mathbb{C}$-algebra which is not induced by an $\mathbb{R}$-algebra, since this would have to be the function field of an elliptic curve over $\mathbb{R}$ which induces the given one over $\mathbb{C}$. But then it would have $j$-invariant in $\mathbb{R}$, a contradiction.