Ring Without Identity - (mZ, +, * )

[Math Amateur]

I am reading Paolo Aluffi's book, Algebra: Chapter 0.

I am currently focused on Chapter III, Section 2: The Category Ring.

I need some help in getting started on Problem 2.15 in this section.

Problem 2.15 at the end of Chapter III, Section 2 reads as follows:View attachment 4482
View attachment 4483I would welcome some help in getting a meaningful start on this problem ...

Peter

 

[Euge]

Hi Peter,

For the first part:

Note that the equation $\varphi(a \bullet b) = \varphi(a) \cdot \varphi(b)$ means $m(a\bullet b) = ma\cdot mb$, or $m(a\bullet b) = m[m(a\cdot b)]$. So $\varphi(a\cdot b) = \varphi(m(a\cdot b))$, which implies $a\bullet b = m(a\cdot b)$ since $\varphi$ is $1-1$. So define $\bullet$ on $\Bbb Z$ such that $a\bullet b := m(a\cdot b)$ for all $a,b\in \Bbb Z$.

For the second part:

To be more explicit, let $\bullet_m$ be the multiplication defined on $\Bbb Z$ such that $a\bullet_m b = m(a\cdot b)$, $a,b\in \Bbb Z$. Suppose $m$ and $n$ are distinct positive integers, and there is a ring isomorphism $T : (\Bbb Z, +,\bullet_m) \to (\Bbb Z,+, \bullet_n)$. We will arrive at a contradiction. If $x := T(1)$, then $T(k) = T(k\cdot 1) = kx$ for all $k\in \Bbb Z$. Hence $T(m) = mx$. On the other hand, $T(m) = T(1\bullet_m 1) = T(1)\bullet_n T(1) = nx^2$. So $mx = nx^2$, i.e., $T(m) = T(nx)$; as $T$ is $1-1$, $m = nx$. Hence, $n$ divides $m$. On the other hand, $T$ has an inverse $S$ such that $S(k) = ky$ for all $k\in \Bbb Z$, where $y = S(1)$. Use the equation $S(T(m)) = m$ to show that $xy = 1$. As a consequence, show that $my = n$. This shows $m$ divides $n$. Since $m$ and $n$ are positive integers and divide one another, they are equal. ($\rightarrow\, \leftarrow$)

 

[Math Amateur]

Hi Peter,

For the first part:

Note that the equation $\varphi(a \bullet b) = \varphi(a) \cdot \varphi(b)$ means $m(a\bullet b) = ma\cdot mb$, or $m(a\bullet b) = m[m(a\cdot b)]$. So $\varphi(a\cdot b) = \varphi(m(a\cdot b))$, which implies $a\bullet b = m(a\cdot b)$ since $\varphi$ is $1-1$. So define $\bullet$ on $\Bbb Z$ such that $a\bullet b := m(a\cdot b)$ for all $a,b\in \Bbb Z$.

For the second part:

To be more explicit, let $\bullet_m$ be the multiplication defined on $\Bbb Z$ such that $a\bullet_m b = m(a\cdot b)$, $a,b\in \Bbb Z$. Suppose $m$ and $n$ are distinct positive integers, and there is a ring isomorphism $T : (\Bbb Z, +,\bullet_m) \to (\Bbb Z,+, \bullet_n)$. We will arrive at a contradiction. If $x := T(1)$, then $T(k) = T(k\cdot 1) = kx$ for all $k\in \Bbb Z$. Hence $T(m) = mx$. On the other hand, $T(m) = T(1\bullet_m 1) = T(1)\bullet_n T(1) = nx^2$. So $mx = nx^2$, i.e., $T(m) = T(nx)$; as $T$ is $1-1$, $m = nx$. Hence, $n$ divides $m$. On the other hand, $T$ has an inverse $S$ such that $S(k) = ky$ for all $k\in \Bbb Z$, where $y = S(1)$. Use the equation $S(T(m)) = m$ to show that $xy = 1$. As a consequence, show that $my = n$. This shows $m$ divides $n$. Since $m$ and $n$ are positive integers and divide one another, they are equal. ($\rightarrow\, \leftarrow$)

Thanks Euge ... but I think i need some further help in order to fully understand what is going on ...

You write:

" For the first part:

Note that the equation $\varphi(a \bullet b) = \varphi(a) \cdot \varphi(b)$ means $m(a\bullet b) = ma\cdot mb$, or $m(a\bullet b) = m[m(a\cdot b)]$. So $\varphi(a\cdot b) = \varphi(m(a\cdot b))$, which implies $a\bullet b = m(a\cdot b)$ since $\varphi$ is $1-1$. So define $\bullet$ on $\Bbb Z$ such that $a\bullet b := m(a\cdot b)$ for all $a,b\in \Bbb Z$. ... ... "

BUT ... ...

... ... we were asked to "use this isomorphism to transfer the structure of 'ring without identity' \(\displaystyle ( m \mathbb{Z} , + , \cdot )\) back onto \(\displaystyle \mathbb{Z}\) ...

I am struggling to understand how what you have said involves using the isomorphism to transfer the structure of 'ring without identity' \(\displaystyle ( m \mathbb{Z} , + , \cdot )\) back onto \(\displaystyle \mathbb{Z}\) ... ... indeed it seems to me that the new multiplication operation \(\displaystyle \bullet\) extends the group homomorphism \(\displaystyle \phi\) (actually isomorphism) into a ring homomorphism ... and that is it ... ? I must be missing something ... certainly not seeing the 'big picture' here ...

Wouldn't using the isomorphism to transfer the structure of 'ring without identity' \(\displaystyle ( m \mathbb{Z} , + , \cdot )\) back onto \(\displaystyle \mathbb{Z}\) involve some ring homomorphism \(\displaystyle \psi : m \mathbb{Z} \rightarrow \mathbb{Z}\)?Hope you can clarify ...

Peter

 

[Euge]

According to the author, he means that one should define a multiplication $\bullet$ on $\Bbb Z$ such that $\phi(a\bullet b) = \phi(a)\cdot \phi(b)$. Indeed, that's what I showed you. If you use $\psi$, which is the inverse of $\phi$, then you'll have $a\bullet b = \psi(\phi(a)\cdot \phi(b))$. That's all there is to it.