4
$\begingroup$

So I am asked to prove or disprove whether the following are isomorphic to one another as rings:

  • $\mathbb{Z}/4\mathbb{Z}$
  • $2\mathbb{Z} /8\mathbb{Z}$
  • $3\mathbb{Z}/12\mathbb{Z}$
  • $\mathbb{R}/4\mathbb{R}$

Now, as far as $\mathbb{R}/4\mathbb{R}$ goes, I am sure that this is not isomorphic to any of the others since $4\mathbb{R}$ contains a unity element and thus the ideal is the entire $\mathbb{R}$.

I know that $\mathbb{Z}/4\mathbb{Z}:=\{0,1,2,3\} \cong \mathbb{Z}_4$, $2\mathbb{Z}/8\mathbb{Z} :=\{0,2,4,6\}$, $3\mathbb{Z}/12\mathbb{Z}:=\{0,3,6,9\} $. Now,because there are only two groups up to isomorphism of order 4, I assume showing that 2 or 3 is isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$ is sufficient to show that it is not isomorphic, but I am unsure no how to proceed.

  • 0
    Can you explain how the fourth isn't infinite? @MattSamuel – 2017-02-28
  • 0
    You actually explained why in your post. $4\mathbb R=\mathbb R$. What is $\mathbb R/\mathbb R$? – 2017-02-28
  • 0
    That would leave us with {0}. – 2017-02-28
  • 0
    Which you would agree is not an infinite ring. – 2017-02-28
  • 0
    Yes, of course. – 2017-02-28
  • 0
    All of the first three rings are cyclic of order 4 as additive groups, so you're not going to show them to be non-isomorphic by the method you suggest. – 2017-02-28
  • 1
    I'll post my conclusion a here. The second isn't ISO to of them because it doesn't have unity meanwhile the first and second do. The first and second are indeed ISO, where the third has 9 as its unity element. Now, the fourth isn't ISO to any because R/4R = R/R = {0}. – 2017-02-28
  • 0
    @ElSpiffy Bingo! – 2017-02-28
  • 0
    You can answer your own question. – 2017-02-28

1 Answers 1

6

Since we have that $4\mathbb{R}$ contains a unity element, we have thus have $\mathbb{R}/4\mathbb{R}=\mathbb{R}/\mathbb{R}=\{0\}$. This is sufficient to say it cannnot be isomorphic to the other three.

Now, $\mathbb{Z}/4\mathbb{Z} \ncong 2\mathbb{Z}/8\mathbb{Z}$ since $\mathbb{Z}/4\mathbb{Z} \cong \mathbb{Z}_4$ and thus has unity element $1$, meanwhile a $2\mathbb{Z}/8\mathbb{Z}$ does not have a unity element. Since isomorphism preserve structural properties, they cannot be isomorphic.

Similiarly, $3\mathbb{Z}/12\mathbb{Z} \ncong 2\mathbb{Z}/8\mathbb{Z}$ by the same argument as above, where $9$ is the unity element in $3\mathbb{Z}/12\mathbb{Z}$.

We then check if $\mathbb{Z}/4\mathbb{Z} \cong 3\mathbb{Z}/12\mathbb{Z}$.

$\phi: 0 \to 0$

$\phi: 1 \to 9$

$\phi: 2 \to 6$

$\phi: 3 \to 3$

Is clearly a bijection and checking the homomorphism property for rings holds, thus we have a ring isomorphism.

Thank you to everyone in the comments for the tips on how to proceed.