- Thread starter
- #1

- Jun 22, 2012

- 2,918

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"We should point out at once that the concept of an R-algebra introduced in 1.9 above occurs very frequently in ring theory, simply because every ring is a Z-algebra. We explain in 1,10 why this is the case."

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Sharp then proceeds as follows:

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"Let \(\displaystyle R \) be a ring. Then the mapping \(\displaystyle F \ : \ \mathbb{Z} \to R \) defined by \(\displaystyle f(n) = n(1_R) \) for all \(\displaystyle n \in \mathbb{Z} \) is a ring homomorphism and, in fact, is the only ring homomorphism from \(\displaystyle \mathbb{Z} \) to \(\displaystyle R \).

Here

\(\displaystyle n(1_R) = 1_R + 1_R + ... \ ... + 1_R \) (n-terms) ... for n > 0

\(\displaystyle n(1_R) = 0_R \) for n = 0

and

\(\displaystyle n(1_R) = (-1_R) + (-1_R) + ... \ ... + (-1_R) \) (n-terms) ... for n < 0

It should be clear from 1.5 (The subring criterion - see attachment) that the intersection of any non-empty family of subrings of a ring R is again a subring of R, This observation leads to the following Lemma (Lemma 1.10 - see attachment, page 6). ... ... ... "

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**My first problem with the above is this:**

Sharp writes: "Let \(\displaystyle R \) be a ring. Then the mapping \(\displaystyle F \ : \ \mathbb{Z} \to R \) defined by \(\displaystyle f(n) = n(1_R) \) for all \(\displaystyle n \in \mathbb{Z} \) is a ring homomorphism and, in fact, is t

**he only ring homomorphism from \(\displaystyle \mathbb{Z} \) to \(\displaystyle R \)**."

But why is this the only ring homomorphism from \(\displaystyle \mathbb{Z} \) to \(\displaystyle R \)?

**My second problem is as follows:**Sharp connects establishing a Z-Algebra (section 1,10 page 6) with establishing the structure of a polynomial ring - but what is the relationship and big picture here?

My third problem is the following:

Dummit and Foote on page 339 define Z-Modules (see attachment) , but seem to structure them slightly differently using a an element a from an abelian group (and not the identity of the group) whereas Sharp uses the multiplicative identity of a ring in establishing a Z-algebra. I presume this is because the abelian groups are being treated as additive and one cannot use the additive identity - can someone please confirm this please - or clarify and explain the links between Z_algebras and Z-modules.

I would really appreciate some help and clarification of these issues.

Peter