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Let $V$ be a $K$-vector space whose algebra of endomorphisms is generated (as a $K$-algebra) by its idempotents.

Is $V$ necessarily finite dimensional?

EDIT (Jul 26 '14) A closely related question:

Is there a field $K$ and a $K$-vector space whose algebra of endomorphisms of is not generated (as a $K$-algebra) by its idempotents?

Pierre-Yves Gaillard
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2 Answers2

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The algebra of endomorphisms of any vector space $V$ is generated by the idempotents.

The finite dimensional case is straightforward, so I will suppose that $V$ is infinite dimensional. Then (assuming the axiom of choice), we have $V\cong W\oplus W$ for some vector space $W$. In fact, we can take $W=V$, as a vector space is determined up to isomorphism by its dimension and, using the fact that $\kappa+\kappa=\kappa$ for infinite cardinals $\kappa$, we have $$ \mathop{dim}(V\oplus V)=\mathop{dim}V+\mathop{dim}V=\mathop{dim}V. $$ So, $V\cong V\oplus V$.

Writing $V\cong W\oplus W$, the endomorphisms of $V$ can be identified with the algebra $M_2(\mathrm{End}(W))$ of $2\times2$ matrices over the endomorphism ring of $W$. We can now express any element of this ring explicitly in terms of idempotents, $$ \begin{align} \left(\matrix{a&b\\c&d}\right)&=\left(\matrix{a-1&0\\0&0}\right)+\left(\matrix{1&b\\0&0}\right)+\left(\matrix{0&0\\c&1}\right)+\left(\matrix{0&0\\0&d-1}\right)\\ \\ &=\left(\matrix{1&a-2\\0&0}\right)\left(\matrix{1&0\\1&0}\right)+\left(\matrix{1&b\\0&0}\right)\\ &+\left(\matrix{0&0\\c&1}\right)+\left(\matrix{0&0\\1&d-2}\right)\left(\matrix{0&1\\0&1}\right). \end{align} $$ [Edit: I replaced the rhs with Pierre-Yves' much simpler variant]. It can be checked that all the matrices on the right hand side of the final equality are idempotents.

George Lowther
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  • Dear George: Thank you very much for your wonderful answer! I just posted a minor variant. – Pierre-Yves Gaillard Aug 02 '14 at 09:01
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    Comment for George, Jonas, and everybody else: George shows that any endomorphism is a linear combination of length 3 words in the idempotents, prompting the question: is 3 the best constant? Letting $c=c(K,V)$ be the best constant, we have the trichotomy: $c=1,c=2$ or $c=3$. Jonas asked (in his first comment to the question): is $c$ always equal to 1? – Pierre-Yves Gaillard Aug 02 '14 at 11:00
  • Thanks. The variant is a bit cleaner - maybe I should replace the rhs of my identity with that? – George Lowther Aug 02 '14 at 11:03
  • I have a slight preference in favour of the replacement you suggest... – Pierre-Yves Gaillard Aug 02 '14 at 11:08
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    Continuation of the previous comment. In fact George shows that any matrix in $M_2(R)$, for any ring $R$, is a $\mathbb Z$-linear combination of length 3 words in the idempotents of $M_2(R)$. So, we can ask the same question as before with $c(R)$ instead of $c(K,V)$. (Here ring means associative ring with 1.) – Pierre-Yves Gaillard Aug 02 '14 at 12:27
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    @Pierre-YvesGaillard: The answer to the last question is no. In the case where $R$ is an integral domain then, if $M\in M_2(R)$ is idempotent then you either have $M$ being diagonal with idempotent components or $\mathrm{trace}(M)=1$. In any case, $\mathrm{trace}(M)$ is a $\mathbb{Z}$-linear combination of idempotents. A positive answer to the question would imply that every element of $R$ is a $\mathbb{Z}$-linear combination of idempotents in $R$ (and, $0,1$ are the only idempotents). This fails in all integral domains except for quotients of $\mathbb{Z}$. – George Lowther Aug 02 '14 at 12:38
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    Thanks! If I understand correctly, you proved that, assuming $R$ is an integral domain, but not a quotient of $\mathbb Z$, we have $c(R)=2$ or $c(R)=3$. - Also, it would perhaps be more natural to suppose that $R$ is an algebra over some commutative ring $K$, and to consider $K$-linear combinations (instead of restricting to the case $K=\mathbb Z$). - I feel that the statement you prove in your answer is much stronger than what was required, so I'm trying to imagine further developments... – Pierre-Yves Gaillard Aug 02 '14 at 13:29
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    It's perhaps worth saying it explicitly: George's identity holds in $M_2(\mathbb Z\langle a,b,c,d\rangle)$, where $a,b,c$ and $d$ are indeterminates, and $\mathbb Z\langle a,b,c,d\rangle$ is the free ring generated by $a,b,c$ and $d$. – Pierre-Yves Gaillard Aug 02 '14 at 15:22
  • Dear George: This is just to tell you that I posted a new answer. – Pierre-Yves Gaillard Aug 03 '14 at 11:09
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    @Pierre-YvesGaillard: Thanks. I see you came up with an even cleaner variant. Mind if I adapt my answer to use it? Also, I can give a proof of Jonas Meyer's question. Every endomorphism of a vector space is a linear combination of idempotents. – George Lowther Aug 04 '14 at 00:27
  • Dear George: Please feel free to adapt your answer to use my variant. I asked Jonas's question separately: http://math.stackexchange.com/q/886864/660. You can guess how anxious I am to see your answer! – Pierre-Yves Gaillard Aug 04 '14 at 05:08
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The purpose of this post is to prove:

(a) If $R$ is a ring (i.e. an associative ring with $1$), then any two by two matrix with entries in $R$ is a $\mathbb Z$-linear combination of products of two idempotents.

Proof: We have

\begin{align*} \begin{pmatrix}a&b\\ c&d\end{pmatrix}&=\begin{pmatrix}1&a-1\\ 0&0\end{pmatrix}\begin{pmatrix}1&0\\ 1&0\end{pmatrix}+\begin{pmatrix}1&b+1\\ 0&0\end{pmatrix}-\begin{pmatrix}1&1\\ 0&0\end{pmatrix}\\ \\ &+\begin{pmatrix}0&0\\ d-1&1\end{pmatrix}\begin{pmatrix}0&1\\ 0&1\end{pmatrix}+\begin{pmatrix}1&0\\ c+1&0\end{pmatrix}-\begin{pmatrix}1&0\\ 1&0\end{pmatrix}. \end{align*}

QED

George Lowther proved:

(b) If $R$ is a ring, then any two by two matrix with entries in $R$ is a $\mathbb Z$-linear combination of products of three idempotents.

George also explained that, in order to answer the question, it sufficed to prove:

(c) If $R$ is a ring, then any two by two matrix with entries in $R$ is a $\mathbb Z$-linear combination of products of idempotents.

Going from (b) to (a) reduces the number of factors from three down to two. This number cannot be reduced further, because, as George observed in a comment, a two by two matrix with entries in $R$ is not in general a $\mathbb Z$-linear combination of idempotents.

Indeed, for trace reasons, $$ \begin{pmatrix}\frac12&0\\ 0&0\end{pmatrix} $$ is not a $\mathbb Z$-linear combination of idempotents of $M_2(\mathbb Q)$.

In fact George shows that, if $R$ is a domain, then the condition that any two by two matrix with entries in $R$ is a $\mathbb Z$-linear combination of idempotents is equivalent to $R$ being a quotient of $\mathbb Z$.

The question about the least number of factors was raised by Jonas Meyer in a comment. Things will become clearer if we generalize slightly the setting.

Let $K$ be a commutative ring and $R$ a $K$-algebra (associative with $1$).

By our previous considerations, any two by two matrix with entries in $R$ is a $K$-linear combination of products of two idempotents, but in general not a $K$-linear combination of idempotents.

Say that $R$ is $K$-nice, or nice over $K$, if any two by two matrix with entries in $R$ is a $K$-linear combination idempotents.

Jonas asked if $\operatorname{End}_K(V)$ (for $V$ a $K$-vector space) is $K$-nice.

One can also ask which rings are nice over their center.

In view of the above proof, $R$ is $K$-nice if and only if any matrix of the form $$ \begin{pmatrix}a&0\\ 0&0\end{pmatrix} $$ is a $K$-linear combination of idempotents.

Pierre-Yves Gaillard
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