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If $R$ is a unital ring and $e$ is an idempotent element, i.e. $e^2=e$, we have that also $(1-e)$ is idempotent. Morover if $e$ is in the centre of $R$ is easy to show that $R \cong R/(e) \ \times R/(1-e)$. The ideal generated by $(e)$ and $(e-1)$ contains $e + (1-e)=1$, thus it is $R$. Whilts if $a$ is in their intersection we have $a=r_1e=r_2(1-e)$, so $0=r_2(1-e)e=r_1e^2=r_1e=a$, so the intersection is trivial (here we have used the fact that $e$ and $e-1$ are in the centre). Therefore $(e)$ and $(1-e)$ are comaximal.

Let $V$ be a vector space on $F$ and $End(V)$ his ring of endomorphisms with, as usual, the pointwise addiction as the sum and the composition as multiplication. Its idempotent elements, called projections, don't need to be in the centre. For example

$A=\begin{pmatrix} 1 & 0 \\ 1 & 0\end{pmatrix}$

clearly $A^2=A$ but

$\begin{pmatrix} 1 & 0 \\ 1 & 0\end{pmatrix}\begin{pmatrix} a & b \\ c & d\end{pmatrix} = \begin{pmatrix} a & b \\ a & b\end{pmatrix}$, meanwhile

$\begin{pmatrix} a & b \\ c & d\end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 0\end{pmatrix}= \begin{pmatrix} a+b & 0 \\ c+d & 0\end{pmatrix}$

But, from the reading I'm doing, the result should hold also in this setting. So I'm looking for a general proof that if $\theta$ is idempotent in $End(V)$ then $(\theta)$ and $(I-\theta)$ are comaximal (or a disprove). Any help would be appreciated, thank you.

Giovanni Barbarani
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1 Answers1

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The statement is wrong. The ring $End(V)$ is simple so the ideals $(\theta)$ and $( 1-\theta)$ are equal to $End(V)$ and are comaximal iff $\theta=0$ or $\theta=1$.

markvs
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  • May you give a proof or a reference that $End(V)$ is simple? Thank you. – Giovanni Barbarani Aug 16 '20 at 22:01
  • Ok, I've found that https://math.stackexchange.com/questions/22629/why-is-the-ring-of-matrices-over-a-field-simple – Giovanni Barbarani Aug 16 '20 at 22:06
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    The standard reference is Jacobson, "Structure of rings", see also https://en.wikipedia.org/wiki/Matrix_ring#Structure – markvs Aug 16 '20 at 22:06
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    The closest to that statement are the Pierce decompositions: https://en.wikipedia.org/wiki/Peirce_decomposition – markvs Aug 16 '20 at 22:11
  • Thanks, for the link. I was trying to interpret such decomposition of $End(V)$ as a vector space as a more general decomposition of $End(V)$ as an $F$-algebra. Now i've realized I was wrong, the product structure is not trivially preserved. – Giovanni Barbarani Aug 16 '20 at 22:19