# Ring which isn't isomorphic to any subring of End(V) for any vector space V

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.

• 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. – Geoffrey Trang May 18 at 19:44

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!

$$V$$ 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 $$R$$.

• Indeed, the ring structure is unnecessary. The direct product cannot be an additive subgroup. – runway44 May 17 at 19:10

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).