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Assume $R$ is an integrally closed ring (in my case: an integral domain), and $r\in\mathbb{N}$.

One can view $\operatorname{End}(R^{n})=\operatorname{Mat}_{n\times n}(R)$ as a submodule of $\operatorname{End}(R^{2n})$ via the inclusion

\begin{equation*} \mu:\left(\begin{matrix}a_{1,1} & ... & a_{1,n}\\ ... & & ...\\ a_{n,1} & ... & a_{n,n}\end{matrix}\right)\longmapsto \left(\begin{matrix}A_{1,1} & ... & A_{1,n}\\ ... & & ...\\A_{n,1} & ... & A_{n,n}\end{matrix}\right)\text{.} \end{equation*}

where $A_{i,j}=\left(\begin{matrix}a_{i,j} & 0\\0 & a_{i,j}\end{matrix}\right)$.

Is $\operatorname{End}(R^{n})$ quadratically closed in $\operatorname{End}(R^{2n})$? By that I mean, is there for any $d\in\operatorname{End}(R^{n})$ an $e\in\operatorname{End}(R^{2n})$ so that $e^{2}=d$? Does for any such $d$ the map $\mu$ extend to a map $\operatorname{End}(R^{n})[t]/(t^{2}-d)\longrightarrow\operatorname{End}(R^{2n})$?

More generally: Is $\operatorname{End}(R^{n})$ closed in $\operatorname{End}(R^{mn})$ with respect to taking $m$th roots?

What happens if I allow for $n=\aleph_{0}$?

The Thin Whistler
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1 Answers1

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I'm afraid I'm a bit skeptical about some of your goals. It is, of course, possible that I have completely misunderstood what you actually want to achieve. Promoting my comments to an answer anyway.

A reordering of the natural basis of $R^{2n}$ from the original linear ordering to the ordering, where the odd-numbered basis elements come first, means that your inclusion homomorphism $\mu$ can be rewritten in the block matrix form as follows: $$\mu: A\mapsto \left(\begin{array}{cc} A&0\\ 0&A \end{array}\right).$$ Because for all $n\times n$ blocks $B$ we have $$ \left(\begin{array}{cc} 0&B\\ I_n&0 \end{array}\right)^2 =\left(\begin{array}{cc} B&0\\ 0&B \end{array}\right) $$ we see that there will, indeed, always be a solution $e\in End(R^{2n})$ of the equation $e^2=\mu(d)$.

However, getting a ring homomorphism from $End(R^n)[t]/\langle t^2-d\rangle$ to $End(R^{2n})$ in the way familiar from the theory of algebraic field extensions is a mirage. The catch is that in the polynomial ring $End(R^n)[t]$ the indeterminate $t$ commutes with all the elements of $End(R^n)$, so we need extra care when substituting something for $t$. The pitfalls were discussed somewhat extensively under the guise of this question, See Arturo Magidin's answer in particular.

For the "evaluate polynomials at $e$" mapping to yield a homomorphism of rings, we need $e$ to commute with all the coefficients of the polynomials. That is, we need $e$ to commute with all of $\mu(End(R^n))$. Or, if you prefer, with all of $End(R^n)$. This obviously won't be the case with my suggested $e$ unless something strange happens. Alas, looking for a different $e$ cannot improve the matter in general. This is because if $e$ commutes with all of $End(R^n)$, so will $e^2=d$. In other words, the mapping $f(t)\mapsto f(e)$, $f(t)\in End(R^n)[t]$, cannot be a ring homomorphism unless $d$ is in the center of $End(R^n)$. That is, $d$ must be a scalar matrix.

Concluding remarks:

  • If $d$ is a scalar matrix, say $d=\alpha I_n$, $\alpha\in R$, then the construction does work, but effectively you will be extending the ring $R$ to $R[\sqrt{\alpha}]$, and the endomorphism ring $End(R^n)$ to a subring of $End(R[\sqrt{\alpha}]^n)$.
  • In the general case the quotient ring $End(R^n)[t]/\langle t^2-d\rangle$ won't even be an extension ring of $End(R^n)$! This is becase for all $a\in End(R^n)$, the element $$a(t^2-d)-(t^2-d)a=ad-da$$ will belong to the ideal that we quotient out. Frankly, I have no idea about what that ideal will look like in general. Matrices of the form $ad-da$ have trace zero, but they can still be non-singular, and we may well end up with the zero ring.
Jyrki Lahtonen
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