# Is the functor that assigns to an algebra its algebraic group of units fully faithful?

Let $$A$$ be a $$d$$-dimensional algebra over a field $$K$$. One can naturally assign to $$A$$ the linear algebraic $$K$$-group $$\mathbf{GL_1}(A)$$ that represents the functor $$B \mapsto (A \otimes_K B)^\times$$ for $$K$$-algebras $$B$$. My question is:

Is the functor $$A \mapsto \mathbf{GL_1}(A)$$ from finite-dimensional $$K$$-algebras to linear algebraic $$K$$-groups fully faithful, i.e. is the natural map $$\hom_{K-\text{Alg}}(A,B) \to \hom_{K-\text{Grp}}(\mathbf{GL_1}(A), \mathbf{GL_1}(B))$$ for finite-dimensional $$K$$-algebras $$A$$ and $$B$$ a bijection?

The $$K$$-algebra $$K$$ doesn't have non-trivial endomorphisms as it is the inital object in $$K-\text{Alg}$$, but the linear algebraic $$K$$-group $$\mathbf{GL_1}(K)$$ does have non-trivial endomorphisms, for example $$x \mapsto x^n$$ for a fixed $$n \in \Bbb N$$, thus the functor is not full.
• So the functor is not full, and consideration of $k[t]$ shows that it is not faithful either. – Kevin Carlson Oct 22 '18 at 17:18