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Problem: Prove that the ring $\mathcal{R}=\prod_{n\geq1}\mathbb{Z}_n$ is not isomorphic to any subring of $\mathrm{End}(V)$ for any vector space $V$.

I think there's something to do with non-commutativity. We know that $\mathrm{End}(V)$ is not commutative. But $\mathcal{R}$ is definitely commutative and of characteristic $0$. Also I think $V$ can't be a vector space of finite dimension.

Thanks in advance!

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  • $\begingroup$ In general, a ring $R$ is not isomorphic to a subring of the endomorphism ring of any vector space over any field if and only if either $R$ has a composite characteristic, or $R$ has characteristic zero with a non-torsion-free additive group. $\endgroup$ – Geoffrey Trang May 18 at 19:44
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Since $\mathcal{R}$ contains an isomorphic copy of $\mathbb{Z}$, the ground field of the vector space $V$ must contain an isomorphic copy of $\mathbb{Q}$ and hence the ground field $k$ of the vector space $V$ must be of characteristic $0$. Now let $\varphi$ be an isomorphism between $\mathcal{R}$ and some subring of $\mathrm{End}(V)$. Let $r=(0,1,0,0,0\ldots)$. Then $2r=r+r=(0,2,0,0,0\ldots)=(0,0,0,0,0\ldots)=0_{\mathcal{R}}$.

Since $\varphi$ is an isomorphism, $2\varphi(r)=\varphi(2r)=\varphi(0)=O_V=2O_V$. Therefore, $\varphi(r)=O_V$. But $r\neq0_{\mathcal{R}}$. Hence a contradiction!

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$V$ is a $k$-vector space. Wlog $k$ is either $\Bbb{Q}$ or $\Bbb{F}_p$.

If it is $\Bbb{F}_p$ then $\forall f\in\mathrm{End}(V)$, $pf=0$.

$R$ contains $\Bbb{Z}$ (send $a\in \Bbb{Z}$ to $(a,a,\ldots) \in R=\prod \Bbb{Z}/n\Bbb{Z}$) so $R\subset \mathrm{End}(V)$ implies that $k=\Bbb{Q}$.

But then $\forall f\in\mathrm{End}(V),\forall a\in \Bbb{Z}\setminus\{0\}, f=0\iff af=0$, which isn't satisfied by $R$.

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    $\begingroup$ Indeed, the ring structure is unnecessary. The direct product cannot be an additive subgroup. $\endgroup$ – runway44 May 17 at 19:10
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For any field $k$ of characteristic zero and any $k$-vector space $V$, $V$ must be a torsion-free abelian group.

Now, suppose that $\mathcal{R}$ is isomorphic to a subring of $\mathrm{End}(V)$ for some $k$-vector space $V$. Then, since $\mathrm{End}(V)$ is a $k$-vector space, it must be a torsion-free abelian group, hence so must $\mathcal{R}$. But $\mathcal{R}$ is clearly not torsion-free ($(0,1,0,0,0,...)$ is an element of order $2$), a contradiction.

If $k$ has a nonzero characteristic $p$, then for any nonzero $k$-vector space $V$, the ring $\mathrm{End}(V)$ must also have characteristic $p$, and so must all of its (unital) subrings. In particular, it cannot have a subring isomorphic to $\mathcal{R}$, which has characteristic zero. And of course, if $V=0$, then $\mathrm{End}(V)=0$ is the zero ring, which obviously has itself as its only subring, so $\mathcal{R}$ still cannot be isomorphic to a subring of $\mathrm{End}(V)$.

The above proof applies more generally for any ring of characteristic zero with a non-torsion-free additive group. Rings with composite characteristics also cannot be isomorphic to subrings of endomorphism rings of vector spaces, of course, but that is it (for a ring $R$, let $V$ be the $\mathbb{Q}$-vector space $R \otimes_{\mathbb{Z}} \mathbb{Q}$ if $R$ has a torsion-free additive group, or the $\mathbb{F}_p$-vector space $R$ if $R$ has characteristic $p$, a prime number).

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