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Let $\mathbb{F}$ be a field, and $\mathbf{V}$ a possibly uncountably generated $\mathbb{F}$-vector space. Let $\mbox{End}_\mathbb{F}(\mathbf{V})$ be the endomorphism ring of $\mathbf{V}$. That the center $\mathbf{Z}(\mbox{End}_\mathbb{F}(\mathbf{V}))=\mathbb{F}$ implies that every maximal commutative subring $R\subset\mbox{End}_\mathbb{F}(\mathbf{V})$ contains $\mathbb{F}$. What can $R$ be? Motivated by finite dimensions I have the following (naive) conjectures.

  1. All such subrings R are isomorphic.

  2. Every choice of a basis $\mathbf{V}\simeq\mathbb{F}^I$ gives rise to a set of commuting $\mathbb{F}$-linear projectors $p_i(v)=v_i$, $i\in I$. The subring generated by $\{p_i\}_{i\in I}$ over $\mathbb{F}$ is a maximal commutative subring.

  3. All maximal commutative subrings $R$ arise in this way.

Question: Are all/some/any of these conjectures true? What would be a readable reference, i.e., free of abstract nonsense? Thank you.

Edit: In view of the comments below, the question needs to be reformulated.

  1. Are all maximal commutative subrings isomorphic?

  2. What is an example of a maximal commutative subring $R$? Thank you.

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    $\begingroup$ IIRC #2 is false and this came up on MO previously. (It's definitely false if the field is not algebraically closed but it's even possible to get a better dimension, I think.) $\endgroup$ Apr 22, 2016 at 18:04
  • $\begingroup$ Thanks for the comment. I would appreciate a reference to the correct answer. $\endgroup$
    – Bedovlat
    Apr 22, 2016 at 18:08
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    $\begingroup$ #3 is also false even if $\mathbb{F}$ is algebraically closed, because $\mathbb{F}[x]$ is a subring of $End(\mathbb{F}^\mathbb{N})$ via $x\mapsto$shift. $\endgroup$ Apr 22, 2016 at 18:08

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Even for a $2$-dimensional vector space (so we're looking at subrings of the ring $M_2(\mathbb{F})$ of $2\times2$ matrices over $\mathbb{F}$) there are nonisomorphic maximal commutative subrings.

Both $$\left\{\begin{pmatrix}x&0\\0&y\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ and $$\left\{\begin{pmatrix}x&y\\0&x\end{pmatrix}: x,y\in\mathbb{F}\right\}$$ are maximal commutative subrings, but they're not isomorphic.

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  • $\begingroup$ Oh, yes, nilpotent elements. Good. $\endgroup$
    – Bedovlat
    Apr 22, 2016 at 19:40

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