Comparing K-linear and F-linear endomorphisms of a K-algebra

Statement

Let $K/F$ be a specified field extension and $A$ a unital associative $K$ -algebra. Use $k\text{-}\mathbf{Alg}_1$ for the category of unital algebras and unit-preserving maps and $k\text{-}\mathbf{Vect}$ for the category of vector spaces and linear maps. After restricting scalars along $F\hookrightarrow K$ , the same underlying functions give inclusions

\begin{array}{ccc} \operatorname{End}_{K\text{-}\mathbf{Alg}_1}(A)&\subseteq&\operatorname{End}_{F\text{-}\mathbf{Alg}_1}(A)\\ \cap&&\cap\\ \operatorname{End}_{K\text{-}\mathbf{Vect}}(A)&\subseteq&\operatorname{End}_{F\text{-}\mathbf{Vect}}(A). \end{array}

The vertical inclusions require multiplication and unit preservation. The horizontal inclusions forget part of the scalar-linearity requirement. The analogous inclusions hold for automorphism groups.

Dimensions in the finite case

If $[K:F]=d<\infty$ and $\dim_K A=m<\infty$ , then $\dim_F A=dm$ , and choices of bases give

\operatorname{End}_K(A)\cong M_m(K),\qquad \operatorname{End}_F(A)\cong M_{dm}(F)

for the vector-space endomorphism algebras. Their automorphism groups are $GL_m(K)$ and $GL_{dm}(F)$ . These formulas do not classify the algebra endomorphisms.

Recovering the extra constraints

An $F$ -linear map is $K$ -linear exactly when it commutes with every scalar multiplication $L_a:x\mapsto ax$ , $a\in K$ . To be a unital algebra map it must also satisfy

f(xy)=f(x)f(y),\qquad f(1)=1.

These are different requirements. They should not be conflated with semilinearity, which allows a field automorphism to twist the scalar action.

Example

Take $K=\mathbb C$ , $F=\mathbb R$ , $A=\mathbb C$ . Conjugation is an $F$ -algebra automorphism but is not $K$ -linear. The real-linear map $z\mapsto2z$ is invertible but preserves neither multiplication nor the unit. The full calculation for this example makes all four collections explicit.