# $\mathbb Z$ is not isomorphic to any proper subring of itself.

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

Is the cardinality main reason for not being isomorphic??

• 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 Tamaroff Jan 4 '15 at 18:52
• 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? – Johanna Jan 4 '15 at 18:52
• 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 4 '15 at 18:54
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.