This is an exercise from a textbook "Algebra: Chapter 0" by Paolo Aluffi.

First, I state necessary facts from the book used in the exercise:

For every abelian group $G, End_{Ab}(G)$( the set of group homomorphisms from $G$ to $G$) is a ring with operations:

$+: (f+g)(a) = f(a) + g(a)$

$.: (fg)(a) = f(g(a))$

and zero $0_{End_{Ab}(G)} = $ trivial map $0: \ \ im0 = 0$, identity $1_{End_{Ab}(G)} = 1_G \in End_{Ab}(G)$

Proposition. $End_{Ab}(\mathbb{Z}) \cong \mathbb{Z}$ as rings.

Let $R$ be a ring. For $r \in R$, define left-multiplication by $r$ by $\lambda_r$. $\lambda_r$ is an endomorphism of the underlying abelian group $(R,+)$.

Proposition. Let $R$ be a ring. Then the function $\lambda: R \to End_{Ab}(R)$ defined by $\lambda(r) = \lambda_r$ is an injective ring homomorphism.

Now, here goes the exercise: it's on the page $138$ in book, numbered $2.16$.

Prove that there is (up to isomorphism) only one structure of ring(with identity) on the abelian group $(\mathbb{Z},+)$. Let $R$ be a ring whose underlying group is $\mathbb{Z}$. Then there is an injective ring homomorphism $\lambda: R \to End_{Ab}(R)$, and the latter is isomorphic to the ring $\mathbb{Z}$. Prove that $\lambda$ is surjective is this case.

So, assume $R$ is some ring, whose underlying abelian group is $(\mathbb{Z}, +)$. We may denote it as $(\mathbb{Z}, + , o , 1_R = k)$ where $o$ is multiplication in $R$, and $k$ is some integer serving and identity in this ring.

We need to prove that every group homomorphism $\phi:(\mathbb{Z},+) \to (\mathbb{Z},+)$ may be defined by formula $\phi(n) = d \ o \ n$ for some integer $d$.

But every group endomorphism $\phi$ of $(\mathbb{Z},+)$ is uniquely determined by the image of $1$. Since $\forall n \in \mathbb{Z} \ \ \phi(n) = \phi(n1) = n\phi(1)$.

And $1_{(\mathbb{Z},+)}$ is determined as $\forall z \in \mathbb{Z} \ \ 1_{(\mathbb{Z},+)}(z) = k(=1_{(\mathbb{Z},+,o)}) \ o \ z = z$.

What can be done next to show $\lambda: (\mathbb{Z}, +, o, k) \to End_{Ab}(\mathbb{Z})$ is surjective?

  • $\begingroup$ In general, for every unital ring $R$, there is a unique unital ring homomorphism $\mathbb{Z}\to R$. $\endgroup$ – egreg Jul 4 '16 at 20:04
  • $\begingroup$ @egreg I know, but how does it help? Have you read my post? $(\mathbb{Z}, +, o)$ is some ring with underlying abelian group $(\mathbb{Z}, +)$, not necessarily the $\mathbb{Z}$ ring. $\endgroup$ – Jxt921 Jul 4 '16 at 20:13
  • $\begingroup$ It doesn't matter; you can't define other ring structures on $\mathbb{Z}$ (with the same addition). The ring homomorphism I told about sends $m$ to the $m$-th multiple of the identity. So the only possible identity is $1$ and the rest easily follows. $\endgroup$ – egreg Jul 4 '16 at 20:16
  • $\begingroup$ @egreg That's what I need to prove! That you can't define other rings structures on $(\mathbb{Z},+)$! And I don't need any proof, I would like the proof in the exercise to be completed first. $\endgroup$ – Jxt921 Jul 4 '16 at 20:21

Suppose $(\mathbb{Z},+,\circ,k)$ is a ring structure ($k$ is the identity). Thus $k\circ x=x$, for every $x\in\mathbb{Z}$.

Suppose $k>0$. Then $k=1+1+\dots+1$, so $$ x=k\circ x=1\circ x+1\circ x+\dots+1\circ x=k(1\circ x) $$ In particular, $x\in k\mathbb{Z}$, for every $x\in\mathbb{Z}$. Therefore $k=1$.

Then for $m>0$ and any $n$, $$ m\circ n=(1+1+\dots+1)\circ n=1\circ n+1\circ n+\dots+1\circ n= n+n+\dots+n=mn $$

If $m<0$, $m\circ n=-((-m)\circ n)=-(-mn)=mn$.

Finish it up, by proving that $k\le0$ is impossible.


$\lambda(k) = \lambda(k \circ k) = \lambda(k)\circ \lambda(k)$ (here the sign on the left is the multiplication in $R$, and the one on the right is functional composition), since we have a ring-homomorphism.

Hence the image of $k$ in $\text{End}_{\text{Ab}}(\Bbb Z)$ is an idempotent endomorphism. Since every endomorphism in $\text{End}_{\text{Ab}}(\Bbb Z)$ is of the form $j \mapsto nj$ for some integer $n$, the fact that the image of $k$ is idempotent means that:

$n^2 = n$, for $\lambda(k) = n\cdot (-)$.

Now $n^2 = n$ is equivalent to: $n(n - 1) = 0$, and since the integers are an integral domain, we have only two choices: $n = 0$, or $n = 1$.

The former case leads to $\lambda$ being the $0$-map, which is certainly not injective. So we must have $n = 1$, which establishes the surjectivity of $\lambda$.

  • $\begingroup$ You need to still "fill in the gaps" by showing that for $r \in R$ that $\lambda(r) = \lambda(k) +\cdots + \lambda(k)$ (for $r$ summands), using the fact that $\lambda$ preserves addition. $\endgroup$ – David Wheeler Jul 4 '16 at 20:31
  • $\begingroup$ So, you proved that $\lambda(k)(1) = 1$. Hence, $\lambda(k) = 1_{\mathbb{Z}}$. But that's have been already known since $\lambda$ is a ring homomorphism, so it sends $1_{(\mathbb{Z}, +, o)}$ to the identity mapping on $G$. How surjecctivity follows from it? Am I missing something? $\endgroup$ – Jxt921 Jul 4 '16 at 22:18
  • $\begingroup$ As an abelian group, $\Bbb Z = \langle 1 \rangle$ (it is cyclic). So if a group homomorphism (which a ring-homomorphism certainly is) maps into $\Bbb Z$ and $1$ is in the image, it is surjective. $\endgroup$ – David Wheeler Jul 4 '16 at 23:07
  • $\begingroup$ Sorry, but I don't understand you. Where do you see a ring homomorphism with a codomain $\mathbb{Z}$? We have $\lambda: (\mathbb{Z}, +, o, k = $identity of the ring ) $\to End_{Ab}(\mathbb{Z})$. $\endgroup$ – Jxt921 Jul 4 '16 at 23:11
  • $\begingroup$ Your first proposition says that $\text{End}_{\text{Ab}}(\Bbb Z) \cong \Bbb Z$ as rings, under the map $(j \mapsto nj) \mapsto n$. Thus post-composition of $\lambda$ with this isomorphism yields a surjective homomorphism into $\Bbb Z$ (in other words, $\text{End}_{\text{Ab}}(\Bbb Z) $ is generated as an abelian group by the identity automorphism). $\endgroup$ – David Wheeler Jul 4 '16 at 23:16

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