# The nilpotence bound in matrix rings

Let $F$ be a field and let $n$ be a positive integer. Does there exist an integer $k$ such that $A^k = 0$ whenever $A\in M_n(F)$ is nilpotent? Here, $M_n(F)$ denotes the ring of $n\times n$ square matrices with entries in $F$.

Such an integer exists. For any nilpotent matrix $A$, we have $A^p = 0$ for some $p$. Therefore, the minimal polynomial of $A$ is of the form $x^{n'}$ for some $n'$. Since the minimal polynomial of $A$ has degree at most $n$, it must be that $A^n = 0$ also. This shows that such an $n$ exists for the ring $M_n(R)$ where $R$ is an integral domain, since $M_n(R)$ is a subring of $M_n(Q(R))$ where $Q(R)$ is the quotient field of $R$.

The ring $M_n(F)$ is the endomorphism ring of the vector space $F^n$. What about infinite-dimensional vector spaces? Then there is no longer an integer $k$ such that $A^k = 0$ for every nilpotent $A$ in the endomorphism ring of an infinite-dimensional vector space.

Here's an example. To keep things simple, let $V$ be a countably infinite-dimensional vector space with basis $v_1,v_2,\dots$. For any positive integer $k$, let $A_k$ be the operator on $V$ defined as follows. Set $A_k(v_1) = 0$ and for $1\leq x \leq k$, set $A_k(v_x) = v_{x_1}$. Finally, set $A_k(v_x) = 0$ if $x > k$. We get a set of operators $A_1,A_2,\dots$ such that
$$A_k^k = 0,$$ but such that $A_k^p\not=0$ for all $p < k$. In general, if $R$ is an associative ring such that there exists an $n$ for which $x^n = 0$ whenever $x\in R$ is nilpotent, then it turns out that $R$ is Dedekind-finite. Dedekind-finite means that for every $x,y\in R$, if $xy = 1$ then $yx = 1$. The condition on nilpotent elements implying Dedekind-finiteness is tricky to prove. Proving that a matrix ring over a field is Dedekind finite can also be done with the classical adjoint matrix: that is, for each $n\times n$ matrix $A$ with coefficients in a commutative ring, there exists a matrix $B$ called the classical adjoint of $A$ such that $AB = BA = \det(A)I_n$, where $I_n$ is the $n\times n$ identity matrix. The existence of the classical adjoint of a matrix immediately implies that matrix rings over commutative rings are Dedekind-finite.

### 1 Comment

• Emily says:

This post represents the spirit of Aleph Zero I've come to know and love over the years.