9
$\begingroup$

According to wikipedia:

If $R$ is a unital commutative ring with an ideal $m$, then $k = R/m$ is a field if and only if $m$ is a maximal ideal. In that case, $R/m$ is known as the residue field. This fact can fail in non-unital rings. For example, $4\mathbb{Z}$ is a maximal ideal in $2\mathbb{Z}$ , but $2\mathbb{Z}/4\mathbb{Z}$ is not a field.

Is $2\mathbb{Z}/4\mathbb{Z}$ NOT a field?! Isn't it isomorphic to $\mathbb{Z_2}$?

I mean every element in $2\mathbb{Z}$ is of the form $2k$ for some $k \in \mathbb{Z}$. Now suppose that I define $\psi: \mathbb{2Z} \to \mathbb{Z}_2$ where $2k \mapsto \bar{0}$ if $k$ is even and $2k \mapsto \bar{1}$ if $k$ is odd.

This map is well-defined because we have only one representation as $2k$ for each element of $2\mathbb{Z}$. This map is surjective because at least $0$ is mapped to $\bar{0}$ and $2$ is mapped to $\bar{1}$.

Is $\psi$ a homomorphism? Well, to me it sounds like it must be. Because the factor of $2$ is playing no role in here and only the parity of $k$ matters for the mapping so I think it is obvious that it is a homomorphism. Even though it can be checked directly by checking all 4 cases separately as well.

So, what is the kernel? Isn't the kernel equal to $4\mathbb{Z}$? Doesn't that mean $2\mathbb{Z}/4\mathbb{Z}$ is isomorphic to $\mathbb{Z}_2$?

It makes sense that these two are isomorphic to me, where am I wrong?

Also, is the ideal correspondence theorem wrong for non-unital rings???

$\endgroup$

2 Answers 2

11
$\begingroup$

Your map is not a ring homomorphism. If I take $a = 2$, $b = 6$, I have $\psi(2) = \psi(6) = 1$, but $\psi(12) = 0$. Likewise, $2\mathbf{Z}/4\mathbf{Z}$ is not an integral ring.

$\endgroup$
5
  • 1
    $\begingroup$ If you allow non-unital rings, you have a nice counterexample here telling you that not every ring is isomorphic to $\mathbf{Z}/2\mathbf{Z}$. :) $\endgroup$
    – fkraiem
    Jan 11, 2014 at 16:30
  • $\begingroup$ No. As an abelian group it is indeed isomorphic, but as a ring it needn't be. $2\mathbb{Z}/4\mathbb{Z}$ has only 2 elements, 0 and $\bar{2}$. And $\bar{2}^2=0$, while in $\mathbb{Z}_2$ it isn't. $\endgroup$ Jan 11, 2014 at 16:30
  • 1
    $\begingroup$ Yeah, I was unaware of the fact that there is more than one ring structure with two elements. Apparently, excluding $\mathbb{Z}_2$, there is at least another ring $R=\{0,1\}$ with the ring multiplication table such that $0.0=0$, $1.0=0$, $0.1=0$, $1.1=0$. I just checked it satisfies the ring axioms. I guess $2\mathbb{Z}/4\mathbb{Z}$ as a ring must be isomorphic to the non-unital ring with two elements that I just mentioned. $\endgroup$
    – user66733
    Jan 11, 2014 at 16:41
  • 1
    $\begingroup$ In general, given an abelian group, the multiplication defined by $ab = 0$ for all $a,b$ gives a non-unital ring, this is what you have here. $\endgroup$
    – fkraiem
    Jan 11, 2014 at 16:44
  • 1
    $\begingroup$ @fkraiem: Thanks. Since earlier I had dealt with only commutative rings and it has become a tradition to consider only unital rings in commutative algebra courses I had never seen a non-unital ring before, so this was a delightful surprise for me today. Thanks for your help. $\endgroup$
    – user66733
    Jan 11, 2014 at 16:59
7
$\begingroup$

Hint $\ 2\Bbb Z/4\Bbb Z\,$ is not a field since $\,2^2 = 0.\,$ Your map $\psi$ is not a ring hom since they must map the nilpotent $\,2\,$ to a nilpotent ($= 0$ in a field), but you have $\,\psi(2)= 1\,$ by definition. Said in further detail $\ 0 = \psi(0) = \psi(2^2)= \psi(2)^2,\, $ so $\,\psi(2) = 0,\,$ being in a field (or domain).

Remark $\ $ Properties like nilpotent, idempotent, invertible etc. that are purely "ring-theoretic" are always preserved by ring homomorphisms. As above, generally this implies constraints on where these elements may be mapped when defining ring homs.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .