For today’s discussion, I will fix a ground field \(k\), and assume \(A\) is an artinian \(k\)-algebra. That is \(A\) is a vector space over \(k\), and I have an unital associative binary operation \((x,y)\mapsto xy\) on \(A\) which is \(k\)-linear in both variables.
I will need the following definitions to formulate today’s problem:
The radical of a subset \(I\) of \(A\) is the set \(\sqrt{I} := \{a\in A\|\ \exists n\in\mathbb{N},\ a^n\in I\}\).
The nil radical \(Nil(A)\) of an \(k\)-algebra \(A\) is the intersection of all of its prime ideals.
Today I want to prove the following result: If \(A\) is artinian then \(Nil(A) = \sqrt{(0)}\).
A note of warning: I do not make any assumptions on the commutativity of \(A\). The proof of this fact when \(A\) is commutative is relatively easy.
An element \(x\in A\) is called nilpotent there exists \(n\in\mathbb{N}\) such that \(x^n=0\). In other words \(\sqrt{(0)}\) is exactly the set of nilpotent elements.
Assume \(x\) is nilpotent. Then there is a natural number \(n\in\mathbb{N}\) such that \(x^n=0\), and since every prime ideal contains \(0\) we can conclude that \(x^n\in P\) for every prime ideal \(P\). Since prime ideals are special in that \(xy\in P\) implies \(x\in P\) or \(y\in P\), we conclude from \(x^n\in P\) that \(x\in P\) for every \(P\). Thus \(x\) is in the nil radical.
Assume now that \(x\) is in the nil radical. Since every maximal ideal is also a prime ideal, we can conclude that \(x\) is in the intersection of all maximal ideals. This intersection is called the Jacobson radical of \(A\) (Notation: \(J(A)\)) and has this nice property that
For every \(x\in J(A)\), the element \(1+x\) is invertible when \(A\) is artinian.
We also have a descending chain of left ideals \[ Ax \supseteq Ax^2 \subseteq Ax^3\supseteq \cdots \] which should stabilize. Then there exists \(n\in\mathbb{N}\) such that \(Ax^n = Ax^{n+1}\). This means there is an element \(a\in A\) such that \[ x^n = ax^{n+1} \quad\implies (ax - 1)x^n = 0 \] But \(x\) is in the nil radical, and therefore, \(x\) is in the Jacobson radical. Therefore, \(ax\) is also in the Jacobson radical which in turn implies \(ax-1\) is invertible. Then we necessarily have that \(x^n=0\). This finishes the proof.
One immediate corollary to the result above is that in an artinian algebra, the radical of a an ideal \(I\) is the intersection of all prime ideals containing \(I\).