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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 $\mathbb{Z}_2$?

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  • $\begingroup$ Did you mean Z4/Z2? $\endgroup$ – Lucio Tanzini Aug 15 at 11:53
  • $\begingroup$ @LucioTanzini $4\Bbb Z \subset 2\Bbb Z \subset \Bbb Z$ $\endgroup$ – ÍgjøgnumMeg Aug 15 at 11:54
  • $\begingroup$ The answer depends on what operation you want on the former structure. Presumably the one induced by the usual operations on the integers. In this case the answer by Wuestenfux is correct. If not, then you really need to specify it . $\endgroup$ – quid Aug 15 at 11:54
  • $\begingroup$ @ÍgjøgnumMeg Oh, I see, sorry $\endgroup$ – Lucio Tanzini Aug 15 at 11:56
  • $\begingroup$ Recall $4\mathbb{Z}\lhd 2\mathbb{Z}$. I meant the quotient ring $2\mathbb{Z}/4\mathbb{Z}$ of the ring $2\mathbb{Z}$. Is it isomorphic to $\mathbb{Z}_2$? $\endgroup$ – mariam Aug 15 at 12:01
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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.

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$\newcommand{\Z}{\mathbb Z}$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.

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