Where my reasoning about module homomorphisms is wrong?

Question

I was playing around trying to find module homomorphisms and I thought I found one in the function $f:\mathbb{Z} \to \mathbb{Z}$ defined by $f(x)=2x$, as $f(x+y)=2(x+y)=2x+2y=f(x)+f(y)$ and $f(rx)=2rx=r(2x)=rf(x)$.

However, if I try to apply the First Isomorphism Theorem, the kernel is $0$ and the image is $2\mathbb{Z}$ which would imply $\mathbb{Z} \cong 2\mathbb{Z}$ which seems problematic as wouldn't that mean $\mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z}$?

I'm sure I'm missing something simple, but I haven't been able to figure out where I went wrong in my reasoning.

Answer 1

Just because two submodules are isomorphic, doesn't mean that their corresponding quotients are isomorphic. Consider, for instance, the Abelian group (and thus $\mathbb Z$-module) $\mathbb Z[X]$ of polynomials with addition as the group operation. It is clearly isomorphic to its own subgroup $\mathbb Z[X^2]$ of polynomials with only even powers. But $\mathbb Z[X]/\mathbb Z[X]$ is clearly trivial, while $\mathbb Z[X]/\mathbb Z[X^2]$ is clearly not trivial, so the quotients are not isomorphic.

Answer 2

$\Bbb Z\cong2\Bbb Z$ is correct. That doesn’t mean you can say $\Bbb Z/\Bbb Z\cong\Bbb Z/2\Bbb Z$ though; the isomorphism is not a literal equality.

Given two rings $R,S$ and $\varphi:R\cong S$, it is true that for any two sided ideal $I$ of $R$ that $R/I\cong S/\varphi(I)$. But in your case we have $R=\Bbb Z$, $S=2\Bbb Z$ and you’re trying to claim $R/I\cong R/\varphi(I)$ which is wrong. But it’s true that $S/\varphi(I)=(2\Bbb Z)/(2\Bbb Z)\cong\Bbb Z/\Bbb Z$.

Answer 3

Why does it seems problematic? Indeed, $\mathbb{Z\cong2Z}$ as $\mathbb{Z}$ modules, but that doesn't imply $\mathbb{\frac{Z}{Z}}\cong\mathbb{\frac{Z}{2Z}}$.