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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!!

Rob Arthan
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Ankit Ghosh
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    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. – Pedro Jan 04 '15 at 18:52
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    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? – user141592 Jan 04 '15 at 18:52
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    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! – Jyrki Lahtonen Jan 04 '15 at 18:54
  • I think it might be helpful to state the definition of "ring" used in your class / book. – Braindead Jan 04 '15 at 18:57
  • @JyrkiLahtonen: In commutative algebra, it's part of the definition. But a lot of book (e.g. Dummit & Foote) doesn't assume this condition when introducing ring, subring... – Krish Jan 04 '15 at 18:57
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    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. – Jyrki Lahtonen Jan 04 '15 at 19:01
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    @JyrkiLahtonen Maybe the teacher wanted to see if his students understood well what a ring isomorphism is, in particular, if they notice that the unit is sent to an unit. – user26857 Jan 04 '15 at 19:10
  • @user26857 More than "understanding", it is a matter of definition. – Timbuc Jan 04 '15 at 19:34
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    @JyrkiLahtonen You're being evil and heretical against the D&F Church of the Last Equations! Repent or burn forever in a nilpotent matrix. – Timbuc Jan 04 '15 at 19:36
  • @Timbuc Some people don't consider $f(1)=1$ as part of the definition. – user26857 Jan 04 '15 at 19:37
  • @user26857 Mopre = more in typonese. And you're making precisely my point: the issue here, beyond or even before understanding, is knowing (and understanding, of course) what the definition given by the teacher/book is. – Timbuc Jan 04 '15 at 19:39
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    @JyrkiLahtonen Watch at Jyrki! Dummit and Foote are known to hand out abstract $\forall ! \iint$ kickings! – Viktor Vaughn Jan 05 '15 at 04:40

1 Answers1

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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.

user26857
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