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!!

  • 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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

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.


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