Let $X$ and $Y$ be abelian groups. Suppose $\text{Hom}(X,Z)\cong \text{Hom}(Y,Z)$ for all abelian groups $Z$. Does it follow that $X \cong Y$?

It has been answered before that this is true if the bijection $\text{Hom}(X,Z)\to \text{Hom}(Y,Z)$ is natural in $Z$. My intuition says that this assumption shouldn't be necessary. Maybe if we choose an extremely large and suitably "generic" group $Z$, then the structure of $\text{Hom}(X,Z)$ will somehow reveal the structure of $X$?

I'm also interested in the answer if "abelian group" is replaced by some other structure, in particular "$R$-module".

  • 1
    $\begingroup$ My three cents: This is true for finitely generated abelian groups, questionable for general abelian groups, and almost certainly false for general $R$-modules. $\endgroup$
    – Slade
    Feb 15 '15 at 1:23

I don't know how to answer this question for abelian groups, but I can answer it over some rings which have a very simple classification of all modules. In particular, suppose $k$ is a field and $X$ and $Y$ are $k$-vector spaces such that $\operatorname{Hom}(X,Z)\cong\operatorname{Hom}(Y,Z)$ (as $k$-vector spaces) for all $k$-vector spaces $Z$. Then it does follow that $X\cong Y$.

To prove this, suppose $X\not\cong Y$. Without loss of generality we may assume $\dim X<\dim Y$; let $\kappa=\dim X$ and $\lambda=\dim Y$. If $\kappa$ is finite the conclusion is trivial by taking $Z=k$, so suppose $\kappa$ is infinite. Let $\mu$ be a strong limit cardinal greater than or equal to $|k|$ and of cofinality $\kappa^+$ and let $Z$ be a $k$-vector space of dimension $\mu$. Then $|Z|=\mu$ so $|\operatorname{Hom}(X,Z)|=\mu^\kappa$ and $|\operatorname{Hom}(Y,Z)|=\mu^\lambda$. But $\mu^\kappa=\mu$ since $\mu$ is strong limit and $\kappa<\operatorname{cf}(\mu)$, whereas $\mu^\lambda\geq\mu^{\kappa^+}=\mu^{\operatorname{cf}(\mu)}>\mu$. Thus $\operatorname{Hom}(X,Z)$ and $\operatorname{Hom}(Y,Z)$ have different cardinalities, and so cannot be isomorphic.

We can generalize this a bit: the answer is also yes for $R$-modules if $R$ is a zero-dimensional principal ideal ring (in particular, for instance, this applies to $R=\mathbb{Z}/(n)$ for any $n>0$). Such a ring is a finite product of local zero-dimensional principal ideal rings, and so we may assume $R$ is local. In that case, $R$ has a nilpotent maximal ideal $m$, every ideal of $R$ is a power of $m$, and every $R$-module is a direct sum of cyclic modules.

So, any $R$-module has the form $$X\cong (R/m)^{\oplus\kappa_1}\oplus (R/m^2)^{\oplus \kappa_2}\oplus\dots\oplus (R/m^n)^{\oplus \kappa_n}$$ for cardinals $\kappa_1,\dots,\kappa_n$, where $n$ is minimal such that $m^n=0$. Now note that $\operatorname{Hom}(R/m^i,R)\cong m^{n-i}R\cong R/m^i$, so if $Z=R^{\oplus\mu}$ then $$\operatorname{Hom}(X,Z)\cong( (R/m)^{\oplus \mu})^{\kappa_1}\oplus ((R/m^2)^{\oplus \mu})^{\kappa_2}\oplus\dots\oplus ((R/m^n)^{\oplus\mu})^{\kappa_n}.$$ When $\mu\geq|R|$, $((R/m^i)^{\oplus\mu})^\kappa$ is isomorphic to $(R/m^i)^{\oplus \mu^\kappa}$ (proof sketch: an $R$-module $N$ is a direct sum of copies of $R/m^i$ iff $\{x\in N:mx=0\}=m^{i-1}N$, and this property is easy to verify for $((R/m^i)^{\oplus\mu})^\kappa$). Thus if you know $\operatorname{Hom}(X,Z)$ up to isomorphism for all $Z$, then you know the cardinals $\mu^{\kappa_i}$ for all $\mu\geq|R|$. As in the previous argument, this uniquely determines the cardinals $\kappa_i$, and so determines $X$ up to isomorphism.

  • $\begingroup$ $|\operatorname{Hom}(X,Y)|=|k|×\dim Y^{\dim X}\neq\dim Y^{\dim X}$,where $k$ denotes the underlying field, so this answer need some modification. $\endgroup$
    – Censi LI
    Jun 12 '20 at 1:35
  • $\begingroup$ That doesn't matter because $\mu\geq|k|$. $\endgroup$ Jun 12 '20 at 1:45
  • $\begingroup$ Surely that's not a big problem, just I think if you consider $\dim(\operatorname{Hom}(X,Z))$ instead of $|\operatorname{Hom}(X,Z)|$, the proof will become more elegant. $\endgroup$
    – Censi LI
    Jun 12 '20 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.