$r\in A$ is in the Jacobson radical iff $\operatorname{tr}(rs)=0\forall s\in A$

Question

Let $A$ be a finite dimensional $k$-algebra for a field $k$ of characteristic zero. Thus, I can regard any element $r\in A$ as a linear function $(r\cdot{})\colon A\to A$ of finite dimensional $k$-vector spaces. I have come across the following statement.

Dickson's criterion: An element $r\in A$ is contained in the Jacobson radical $J(A)$ if and only if $\operatorname{tr}(rs)=0$ for all $s\in A$.

However, I have failed to find a proof, either by myself or by looking on the internet. I am even not sure if the statement is correct this way. Any precisition, proof and/or reference is appreciated.

rschwieb · Accepted Answer · 2019-05-24T15:17:31.717

4

Interesting: I haven't seen this before now.

I think you want to make use of this fact that

For a field of zero characteristic, if the powers of a matrix all have trace zero, then the matrix is nilpotent.

I would have liked to provide a concrete reference for this, but I haven't found one. All sorts of references prove this for $\mathbb R$ and $\mathbb C$, but I'm fairly sure it holds for fields of characteristic $0$ in general. I'm pretty sure the approach given at this solution indicates that it is so.

So in your case, given $s\in A$, $0=\mathrm{tr}(rs)=\mathrm{tr}(rsrs)=\cdots$ so that $rs$ is nilpotent. This means $rA$ is a nil right ideal, and the Jacobson radical contains all nil right ideals.

Conversely, since the Jacobson radical of an Artinian ring is a nilpotent ideal, anything of the form $rs$ with $r\in J(A)$ and $s\in A$ will be nilpotent as well, and as is well known nilpotent transformations have trace $0$.

edited May 24 '19 at 15:17

answered May 02 '19 at 16:36

rschwieb

153,510

2

See Corollary 4.1 (b) in my note The trace Cayley-Hamilton theorem for your fact, stated for arbitrary commutative $\mathbb{Q}$-algebras (but keep in mind that you need the field to have characteristic $0$, or at least not dividing $n!$, where $n$ is the size of the matrix). – darij grinberg May 02 '19 at 17:32
Also note that you have only proven one direction of the criterion. – darij grinberg May 02 '19 at 17:33
I am wondering to what extent this all can be generalized beyond fields. Let $\mathbb{K}$ be a commutative ring in which $1, 2, \ldots, n$ are invertible, and let $A$ be a $\mathbb{K}$-algebra that is a finite free $\mathbb{K}$-module. Is it true that $J\left(A\right) = \left{a \in A \mid \operatorname{Tr}\left(as\right) \in J\left(\mathbb{K}\right) \text{ for all } s \in A\right}$ ? – darij grinberg May 02 '19 at 17:38
@darijgrinberg I went ahead and put in a note about the trivial direction. Thanks. And thanks for the citation! – rschwieb May 02 '19 at 17:53
And wait.. what? I put "positive characteristic"? That's the opposite of what I intended. – rschwieb May 02 '19 at 17:55
As $tr(ab)=tr(ba)$ I wonder if looking at $A/I\cong k[x]/(f(x))$ with $I$ the two-sided ideal generated by the $ab-ba$ can help – reuns May 02 '19 at 17:55
The question asked in my third comment here now has its own MathOverflow thread. – darij grinberg May 26 '19 at 13:24

$r\in A$ is in the Jacobson radical iff $\operatorname{tr}(rs)=0\forall s\in A$

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