The Jacobson radical of a finite-dimensional algebra is the intersection of its maximal left ideals

Let $A$ be a finite-dimensional algebra over a field. Its Jacobson radical is

$J(A):=\bigcap \{L<A:L\text{ is a maximal left ideal of }A\}\mathrm{.}$

When $A\mathrm{\ne }0$, maximal proper left ideals exist without a choice principle: the proper left ideal $0$ exists, and among dimensions of proper left ideals choose the largest integer. An ideal of that dimension is maximal because strict containment strictly increases vector-space dimension. On finite-dimensional algebras $J(A)$ is two-sided: for each simple left $A$-module $S$ and each $0\mathrm{\ne }s\in S$, the kernel of the surjection $A\to S$, $a\mapsto as$, is a maximal left ideal. Hence an element in every maximal left ideal annihilates every simple left module. Conversely, an element annihilating every simple left module lies in each maximal left ideal by applying it to $A\mathrm{/}L$ and $1+L$. Thus $J(A)$ is the intersection of the two-sided annihilators of all simple left modules, so $A\mathrm{/}J(A)$ is an algebra. This radical agrees with the usual right-sided definition.