Are the rings $2\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}_2$ isomorphic?

Question

The rings $2\mathbb{Z}/4\mathbb{Z}\cong \{\bar{0},\bar{2}\}$ and $\mathbb{Z}_2=\{\bar{0},\bar{1}\}$ have order two. So, are they isomorphic?

However, $\mathbb{Z}_2$ is a field but $\{\bar{0},\bar{2}\}$ looks not to be!

To be more specific, recall $4\mathbb{Z}\lhd 2\mathbb{Z}$, and I mean the quotient ring $2\mathbb{Z}/4\mathbb{Z}$ of the ring $2\mathbb{Z}$.

Is it isomorphic to $\mathbb{Z}_2$?

Answer 1

Well, ${\Bbb Z}_2$ is a field with $\bar 1\cdot\bar 1 = \bar 1$.

But in $2{\Bbb Z}/4{\Bbb Z}$, we have $\bar 2\cdot \bar 2 = \overline{2\cdot 2} = \bar 4 = \bar 0$ and so this ring has zero divisors.

Thus the rings cannot be isomorphic.

Answer 2

$\newcommand{\Z}{\mathbb Z}$Note that for $x,y\in2\Z/4\Z$ we have $xy = 0$, i.e. the multiplicative structure is trivial. In particular, this ring doesn't even have unit.

But what are multiplicative maps $f\colon \Z/2\Z\to 2\Z/4\Z$? Well, $$f(x) = f(x\cdot(1+2\Z)) = f(x)f(1+2\Z) = 0,\ \forall x\in \Z/2\Z,$$

and thus the only multiplicative map is trivial.

$\Z/2\Z$ is isomorphic to $2\Z/4\Z$ as additive group, though, the isomorphism being $x+2\Z\mapsto 2x + 4\Z$. This is actually the only nontrivial additive homomorphism. This map is not multiplicative, so this is another way to see that there is no ring isomorphism between those.