Prove $2\mathbb{Z}$ and $3\mathbb{Z}$ are not isomorphic as rings.

Question

I want to prove the ring of even integers $2\mathbb{Z}$ is not isomorphic to $3\mathbb{Z}$. I do this by contradiction, assume $\phi$ is such an isomorphism from $2\mathbb{Z}$ to $3\mathbb{Z}$. Observe this equality:

$$\phi(2)+\phi(2)=\phi(2+2)=\phi(2\cdot2)=\phi(2)\cdot\phi(2)$$

Let $x=\phi(2)$ and we have the equation $x^2-2x=0$ in $3\mathbb{Z}$. This factors into $x(x-2)=0$.

$\phi(2)=0$ is not possible because any isomorphism will have $\phi(0)=0$.

Thus we are left with $x-2=0$ which has no solution in $3\mathbb{Z}$.

Question. Is this a valid argument? What does the $2$ symbol in the last equation denote, if not an element of $3\mathbb{Z}$?

Answer 1

$x^2 - 2x = x(x-2)$ is not a valid factorization in $3\mathbb{Z}$ precisely because $2 \notin 3\mathbb{Z}$.

But you can just directly conclude that $x^2 - 2x = 0$ has no solutions in $3\mathbb{Z}$.

Namely, if $x = 3k$ then

$$9k^2 = x^2 = 2x = 6k \implies 9k= 6$$

which is a contradiction.

Answer 2

Suppose $\phi:2\mathbb{Z}\to3\mathbb{Z}$ were an isomorphism. Then $$2\phi(2) = \phi(2) + \phi(2) = \phi(4) = \phi(2)\phi(2)\ .$$ This is only possible if either $\phi(2) = 0$, in which case $\phi$ cannot be an isomorphism, or if $\phi(2) = 2\notin3\mathbb{Z}$.

Answer 3

Seeing everything as sub(non unital)rings of $\mathbb{Z}$ makes your argument work.

But you can also write your equation as $x^2=x+x$, and plug in $3k$, and see that it has no solution.

Answer 4

On your question of what does the $2$ denote. In a field or ring with unity, it is common to define $2 = 1 + 1$, $3 = 2 + 1$ etc. Many of the properties of these elements are familiar but you must avoid assuming that they are necessarily non-zero. Even in a field, it is quite possible that $1 + . . . + 1 = 0$. If you want to know more, look up the characteristic of a field or ring.

Even for a ring without unity, we can define $2x = x + x$, $3x = x + x + x$, etc even though we cannot identify $2$ with an element of the ring. This is no more strange than $x^2 = x \times x$. $x^3 = x \times x\times x$, etc. The two are analogous. For Abelian groups, it is common to write the operation as addition in which case $nx$ would be used in favour of $x^n$.

Edit: mechanodroid makes the good point that your rings do not have unities hence my first definition of $2$ is not applicable. The second is but that is not what you have. So, this is sometimes useful but not in this case.

Answer 5

Your argument is correct until $x^2 - 2x = 0$ which suffices to complete the proof. Here is another way to see this:

Your ring isomorphism $\phi$ is an isomorphism of abelian groups which are both infinite cyclic. Hence the generator $2$ of $2\mathbb{Z}$ is mapped to a generator of $3\mathbb{Z}$, i.e. $\phi(2) = \pm3$. This implies $\pm 6 = \phi(2) + \phi(2) = \phi(2 + 2) = \phi(2 \cdot 2) = \phi(2) \cdot \phi(2) = (\pm 3) \cdot (\pm3) = 9$, a contradiction.

Answer 6

Let us consider the following general problem:

Determine all ring homomorphisms $\phi : n\mathbb{Z} \to m\mathbb{Z}$, where $n,m \in \mathbb{N}$.

All $n\mathbb{Z}$ are subrings of $\mathbb{Z}$ (non-unital if $n > 1$).

So let $\phi$ be a ring homomorphism. We obtain

$$n\phi(n) = \phi(n) + ... + \phi(n) = \phi(n + ... + n) = \phi(n \cdot n) = \phi(n) \cdot \phi(n) . $$

This is an equation in $\mathbb{Z}$; it is equivalent to

$$\phi(n)^2 - n\phi(n) = \phi(n) \cdot (\phi(n) - n) = 0 .$$

It has the two solutions $\phi(n) = 0$ and $\phi(n) = n$.

In case $\phi(n) = 0$ we obtain $\phi = 0$ (because the generator $n$ of the cyclig group $n\mathbb{Z}$ is mapped to $0$). Note that $0$ is always a ring homomorphism between non-unital rings.

The case $\phi(n) = n$ is only possible when $n \in m\mathbb{Z}$, i.e. when $m$ divides $n$ which is true if and only if $n\mathbb{Z}$ is a subring of $m\mathbb{Z}$. We obtain for all $r \in \mathbb{Z}$

$$\phi(rn) = r\phi(n) = rn$$

This means that $\phi$ is the inclusion of subrings $n\mathbb{Z} \subset m\mathbb{Z} \subset \mathbb{Z}$.

In other words, there exists at most one non-zero ring homomorphism $\phi$ which is the subring inclusion if $n\mathbb{Z} \subset m\mathbb{Z}$.

We moreover see that $n\mathbb{Z}$ and $m\mathbb{Z}$ are isomorphic rings if and only if $n = m$. This answers your question ($n = 2$,$m = 3$).

Let us finally consider the above equation $x^2 - nx = 0$ in $m\mathbb{Z}$. The term $nx$ is the sum of $n$ summands $x$. Only when $n \in m\mathbb{Z}$ we can write $nx = n \cdot x$ as a product in $m\mathbb{Z}$. Therefore in general the expression $x^2 - nx$ cannot be regarded as a polynomial in $m\mathbb{Z}[x]$. In particular it does not make sense to think about a polynomial factorization $x^2 - nx = x \cdot (x - n)$ in $m\mathbb{Z}[x]$. All that works precisely when $n$ divides $m$. Note that in our above calculation of $\phi(n)$ we used factorization in $\mathbb{Z}$, but we could also factorize the polynomial $x^2 - n \cdot x$ in $\mathbb{Z}[x]$.

Answer 7

The argument is valid, because $2\mathbb{Z}$ and $3\mathbb{Z}$ are subrings of $\mathbb{Z}$ and nothing prevents you from using this larger ring for doing computations in.

A simpler argument is that the equation $x+x=6$ has solution in $3\mathbb{Z}$, but no solution in $2\mathbb{Z}$.