2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

$\lambda(k) = \lambda(k \circ k) = \lambda(k)\circ \lambda(k)$ (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 $\lambda$.

$\endgroup$
  • $\begingroup$ 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

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.