Endomorphisms of Q(i) as a rational vector space and algebra

For $A=\mathbb Q(i)$ , the basis $(1,i)$ identifies its rational-linear endomorphism algebra and automorphism group with

\operatorname{End}_{\mathbb Q\text{-}\mathbf{Vect}}(A)\cong M_2(\mathbb Q), \qquad \operatorname{Aut}_{\mathbb Q\text{-}\mathbf{Vect}}(A)\cong GL_2(\mathbb Q).

In the unit-preserving rational algebra category, however,

\operatorname{End}_{\mathbb Q\text{-}\mathbf{Alg}_1}(A) =\operatorname{Aut}_{\mathbb Q\text{-}\mathbf{Alg}_1}(A) =\{\operatorname{id},\;a+bi\mapsto a-bi\}.

The latter automorphism group is cyclic of order two.

Direct proof

A rational-linear map may send the two basis vectors anywhere, giving any $2\times2$ rational matrix. It is invertible exactly when its determinant is nonzero.

A unital rational algebra map fixes every rational number. Its image of $i$ must be a root of $X^2+1$ in $\mathbb Q(i)$ , so is $i$ or $-i$ . Both choices extend to the displayed automorphisms. If units are not required, the zero map is the only additional algebra endomorphism: a nonzero map sends $1$ to a nonzero idempotent in a field, hence to $1$ .

Scalar embeddings as matrices

Multiplication by $a+bi$ is the rational-linear matrix

\begin{pmatrix}a&-b\\b&a\end{pmatrix}.

These matrices form the endomorphisms linear over $\mathbb Q(i)$ itself, a proper subalgebra of $M_2(\mathbb Q)$ . Requiring preservation of the algebra unit leaves only multiplication by $1$ in that scalar-linear family.