Let $V$ be an infinite-dimensional $K$-vector space, $E=\operatorname{End}_K(V)$ and $I = \{f \in E:\dim_Kf(V) < \infty\}$.
I want to prove that $I$ is an ideal of $E$ and $K\mathrm{id}_V + I$ is a subring of $E$.

My idea:

Let's first prove that $I$ is an ideal of $E$. Let $f$ be an element of $I$ and $g$ an element of $E$.

Then, $dim((f \circ g)(V)) \le dim(f(g(V))) \le dim(f(V)) \lt \infty$ and $dim((g \circ f)(V)) \le dim(g(f(V))) \lt \infty$. This proves that $I$ is an ideal of $E$.

Let' s prove next that $Kid_V + I$ is a subring of $E$. Let $k_1id_V + f_1$ and $k_2id_V+f_2$ be elements of $I$.

Then, $(k_1id_V+f_1) \circ (k_2id_V+f_2) = k_1k_2id_V + k_1f_2 + k_2f_1 + f_1 \circ f_2$.

Futhermore, $dim(k_1f_2 + k_2f_1 + f_1 \circ f_2)\le dim(k_1f_2) + dim(k_2f_1) + dim(f_1 \circ f_2) \lt \infty$.

Besides, $id_V \in I$ and $-(k_1id_V+f_1) \in I$ as well as $0 \in I$.

This proves that $I$ is a subring.

Is this correct?

By the way, does someone know if $V$ is a simple module?

  • $\begingroup$ The line that begins with "Furthermore" is dispensable if you already proved $\;I\;$ is an ideal. Yet $\;Id_V\notin I\;$ as $\;Id_V(V)= V\;$ is infinite dimensional. $\endgroup$
    – DonAntonio
    Jul 19, 2016 at 18:38
  • 1
    $\begingroup$ If $R$ is a ring with a subring $S$ and an ideal $I$, then $S+I$ is a subring (no matter what the ring is). $\endgroup$
    – egreg
    Sep 2, 2016 at 21:20

1 Answer 1


You needn't be checking dimensions when considering $Kid_V+I$. You say $id_V\in I$, but this is obviously false: the image of $id_V$ is all of $V$, which is infinite dimensional by definition. This new ring properly contains $I$, and so it necessarily contains things with infinite dimensional images.

It is enough to show that the set of things of the form $\{k\cdot1_V+ i\mid k\in K, i\in I\}$ is closed under addition and multiplication, and that is easy enough. The only thing in contention really is whether or not $(k\cdot 1_V)\cdot i$ and $i\cdot (k\cdot 1_V)$ have finite dimensional images, but knowing that $I$ is an ideal makes this obvious.

Yes, $V$ is a simple module (when acted upon by linear transformations from this ring in the obvious way.) Given $v,w$ nonzero elements of $V$, you can easily construct a linear transformation that takes one to the other. This amounts to the module being simple.

  • $\begingroup$ As I' m currently dealing with artinian and noetherian rings, I' d be curious about whether the subring $I$ is artinian or noetherian. Do you know that? $\endgroup$
    – Peter123
    Jul 19, 2016 at 19:27
  • $\begingroup$ @Peter123 My gut says "no" to both, but I haven't thought of a way to do it yet. One thing you have going for you is that $I$ is von Neumann regular (that is, for every $a$, there exists $b\in I$, $a=aba$.) Also it is a simple ring (since ideals of $I$ are ideals of $R$ too in a VNR ring.) For rings with identity that would be a great help, but $I$ does not have an identity. $\endgroup$
    – rschwieb
    Jul 20, 2016 at 15:16

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