2
$\begingroup$

Show that the ring $ \mathbb Z$ is not isomorphic to any proper subring of itself.

Is the cardinality main reason for not being isomorphic??

Please Help!!

$\endgroup$
12
  • 5
    $\begingroup$ Are you asking for subrings to share the unit of the original ring? In such case $\Bbb Z$ may only have $0$ has the only subring, which other people might not even call a ring. $\endgroup$ – Pedro Tamaroff Jan 4 '15 at 18:52
  • 2
    $\begingroup$ It's not cardinality. Think about the subrings $n\mathbb{Z}$. They are all infinite too. Can you build an isomorphism to any of them? $\endgroup$ – Johanna Jan 4 '15 at 18:52
  • 4
    $\begingroup$ I belong to the church believing that all rings have a multiplicative neutral element. Your teacher apparently doesn't. But, humoring them, which subrings of $\Bbb{Z}$ have a neutral element? Isn't having a neutral element a property preserved by isomorphisms? According to anyone's definition! $\endgroup$ – Jyrki Lahtonen Jan 4 '15 at 18:54
  • 2
    $\begingroup$ Sigh. @Krish: The topic has been discussed THOROUGHLY on our site. For example here. Hmm, this is a better fit. I did say believing :-). Dummit & Foote do not have authority over all practitioners. $\endgroup$ – Jyrki Lahtonen Jan 4 '15 at 19:01
  • 3
    $\begingroup$ @JyrkiLahtonen You're being evil and heretical against the D&F Church of the Last Equations! Repent or burn forever in a nilpotent matrix. $\endgroup$ – Timbuc Jan 4 '15 at 19:36
7
$\begingroup$

Let $f:\mathbb Z\to A$ be a ring isomorphism, where $A\subsetneq\mathbb Z$ is a subring. Then $f(1)=a\in A$, and from $f(1)^2=f(1)$ we get $a=1$ or $a=0$. In the first case $A=\mathbb Z$, a contradiction, while in the second $f$ isn't injective (recall that $f(0)=0$).

For short, $\mathbb Z$ has no proper unitary subrings.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.