Absolutely simple objects

Let $k$ be a field and let $\mathcal{C}$ be a $k$-linear category (k-linear categories and k-linear functors). Since composition in $\mathcal{C}$ is $k$-bilinear, ${\mathrm{End}}_{\mathcal{C}}(X)$ is a $k$-algebra with unit ${\mathrm{id}}_X$ for every object $X$.

An object $X$ of $\mathcal{C}$ is absolutely simple when the $k$-algebra ${\mathrm{End}}_{\mathcal{C}}(X)$ is one-dimensional over $k$, that is,

${\mathrm{End}}_{\mathcal{C}}(X)=k\cdot {\mathrm{id}}_X,{\mathrm{id}}_X\mathrm{\ne }0.$

Equivalently, the evaluation map

$k⟶{\mathrm{End}}_{\mathcal{C}}(X),\lambda ⟼\lambda {\mathrm{id}}_X,$

is an isomorphism of $k$-algebras. These two formulations agree: the evaluation map is $k$-linear and multiplicative with $1\mapsto {\mathrm{id}}_X$, and it is injective because $\lambda {\mathrm{id}}_X=0$ with ${\mathrm{id}}_X\mathrm{\ne }0$ forces $\lambda =0$; surjectivity is exactly ${\mathrm{End}}_{\mathcal{C}}(X)=k\cdot {\mathrm{id}}_X$. In particular a nonzero endomorphism space $k\cdot {\mathrm{id}}_X$ has dimension one, so an absolutely simple object is not a zero object; if $\mathcal{C}$ has a zero object then $X\mathrm{\ne }0$ automatically.

In a $k$-linear abelian category, a simple object with $\mathrm{End}(X)=k$ satisfies this condition (Simple object). The converse need not hold: the representation $k\stackrel{1}{\to }k$ of the two-vertex quiver has only scalar endomorphisms but has the proper nonzero subrepresentation $0\to k$. Thus the definition records the scalar endomorphism condition without asserting simplicity in an arbitrary abelian category. Every simple object of a locally finite $k$-linear abelian category over an algebraically closed field $k$ is absolutely simple. Indeed, a nonzero endomorphism of a simple object has zero kernel and full image, hence is an isomorphism; its endomorphism algebra $D$ is therefore a division algebra. Local finiteness makes $D$ finite-dimensional (Locally finite k-linear abelian categories). For $a\in D$, a polynomial over $k$ annihilates $a$ by linear dependence of its powers; it splits into linear factors, and a division algebra has no zero divisors, so one factor $a-\lambda {\mathrm{id}}_X$ vanishes. Thus $D=k$. This definition fixes no semisimplicity and no algebraic-closedness hypothesis; it records them only as the standard situation in which absolute simplicity is automatic.

If $X$ is absolutely simple, every automorphism of $X$ is a scalar $\lambda {\mathrm{id}}_X$ with $\lambda \in k^{\times }$: it is an invertible endomorphism, hence by absolute simplicity equals some $\lambda {\mathrm{id}}_X$, and $\lambda$ is invertible with inverse ${\lambda }^{-1}$ because $\lambda {\mathrm{id}}_X$ is invertible.