Let $A$ be a finite-dimensional unital commutative associative algebra over a field $K$ of characteristic $0$. Is it true that for any derivation $D$ of $A$ we have $D(A) \subseteq J(A)$ where $J(A)$ is the Jacobson radical of $A$?

Please help me with this question or give some references.


1 Answer 1


I suppose you want $D$ to be a $K$-derivation, otherwise there are obvious counter-examples with $A=K$. Then the answer is yes, but for somewhat stupid reasons. First of all, $A$ is a product of a finite number of local rings, and $D$ respects this decomposition: if $e\in A$ is idempotent, from $e^2=e$ one gets $(2e-1)De=0$, hence $De=0$, and therefore $D(Ae)\subset Ae$. Thus one can assume that $A$ is local, with maximal ideal $\mathfrak{m}$. Let $x\in\mathfrak{m}$; there exists an integer $n\geq 1$ such that $x^n=0$ but $x^{n-1}\neq 0$. Then $nx^{n-1}D(x)=0$, which means that $D(x)$ cannot be invertible, hence $D(x)\in\mathfrak{m}$.

Edit : As observed by @Keith Kearnes, the argument is completed as follows: we have shown $D(\mathfrak{m})\subset \mathfrak{m}$, so that $D$ induces a $K$-derivation of $A/\mathfrak{m}$. But $A/\mathfrak{m}$ is a finite extension of $K$, so any $K$-derivation of $A/\mathfrak{m}$ is zero. But this means $D(A)\subset \mathfrak{m}$.

  • $\begingroup$ This doesn't seem so stupid to me coming from the background of Banach algebras: it's an old result of Singer and Wermer that if A is a unital commutative Banach algebra over the complex field and $D:A\to A$ is a continuous complex-linear derivation, then $D(A)\subseteq J(A)$. $\endgroup$
    – Yemon Choi
    Sep 2, 2018 at 3:29
  • 4
    $\begingroup$ The argument is not yet complete: you show that $D(J(A))\subseteq J(A)$, but not that $D(A)\subseteq J(A)$. Continue (in the local case): if $x\in A-J(A)$ does not satisfy $D(x)\in J(A)$, differentiate the minimal polynomial of $x$ over $K$ modulo $J(A)$ to get a contradiction. $\endgroup$ Sep 2, 2018 at 5:50

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