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This is a problem from gtm 167 Field and Galois Theory.

Let R be a commutative ring with identity.The prime subring of R is the intersection of all subrings of R.Show that this intersection is a subring of R that is contained inside all subrings of R.Moreover,show that the prime subring of R is equal to $\{n\cdot1:n\in\mathbb{Z}\}$,where 1 is the multiplicative identity of R

The first part of this problem is easy.What puzzled me is the second part.I think under this definition, the prime subring of R should be $0$.Even if we only consider the inersection of non-zero subring of R,I think the outcome won't be $\{n\cdot1:n\in\mathbb{Z}\}$.Since $\{2n\cdot1:n\in\mathbb{Z}\}$ is also a subring of R.This is a controdiction.I want to know where I was wrong

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  • $\begingroup$ Is $2\mathbb{Z}$ considered a subring of $\mathbb{Z}$? It does not contain the identity of $\mathbb{Z}$. $\endgroup$ – Lozenges May 31 '19 at 5:29
  • $\begingroup$ @Lozenges It will make sense if we ask the subring of R should contain the identity of R. But I remenber the definition of subring don't require that... $\endgroup$ – Y.Wayne May 31 '19 at 9:23
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This definition of “prime subring” is inconsistent with the definition of subring that does not require shared unity.

It could simply be a mistake, yes, or maybe there is more to the context than you are letting on. From scanning a copy it seems there are some small mistakes in the book of this sort.

I can’t actually locate an explicit definition of “subring” in the book... did you? If not, I would not get hung up on this and I would advise using the stronger definition.

By the way, the author does explicitly say all rings in the book should have identity, and that would make subring a necessarily have identity, eliminating one of your examples.

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  • $\begingroup$ Thanks a lot. I find the author's assumption about ring in the appendix. $\endgroup$ – Y.Wayne May 31 '19 at 10:27

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