9
$\begingroup$

I think this definition is wrong, because nothing guarantees that the subring is closed to additive inverses.

Thanks

$\endgroup$
  • 2
    $\begingroup$ I don't think I quite understand why you think his definition is wrong... $\endgroup$ – rfauffar Sep 24 '13 at 14:54
  • 6
    $\begingroup$ @RobertAuffarth: The definition seems to imply that $\mathbb N$ is a "subring" of $\mathbb Z$. $\endgroup$ – Henning Makholm Sep 24 '13 at 14:57
  • 1
    $\begingroup$ Oh! Now I understand, thanks. $\endgroup$ – rfauffar Sep 24 '13 at 15:00
  • $\begingroup$ Yep, seems like a problem. What edition of the book are you looking at? A "hack" fix would be to change it to: "and contains the additive inverse of the identity element of $A$." But better to just assert that $S$ is an additive subgroup. $\endgroup$ – Thomas Andrews Sep 24 '13 at 15:01
  • $\begingroup$ @ThomasAndrews I think it's the first one from 1969. $\endgroup$ – user42912 Sep 24 '13 at 15:06
12
$\begingroup$

Yes, it is a typo. In Russian translation of the book it is written: "... if $S$ is an additive subgroup, closed under multiplication ..."

$\endgroup$
  • 4
    $\begingroup$ Clever guys the Russians. The mistake went through in the Spanish translation :) $\endgroup$ – Adrián Barquero Sep 24 '13 at 15:05
  • 7
    $\begingroup$ @Adrián Barquero No, not all Russians are clever. :-) The book was translated by Yu.Manin. $\endgroup$ – Boris Novikov Sep 24 '13 at 15:11
1
$\begingroup$

I cannot believe that the definition of a subring in Atiyah-Macdonald is wrong or incomplete. I still try to find a way to read the definition so that it becomes true ;-).

By the way, instead of adding "if $S$ is an additive subgroup ...", it suffices to demand that $-1 \in S$. Because then we have $-x = (-1) \cdot x \in S$ for all $x \in S$.

By the way, even then this won't be the "correct" definition: A subring is not just a subset with properties. It is a subset with extra structure, namely a ring, and properties. A quite abstract definition would be: A subring of $R$ is a subobject of $R$ in the category of rings. Equivalently, and more explicit: A subring $S$ of $R$ is a ring (!) together with a homomorphism of rings $S \to R$, which is just an inclusion on the underlying sets. But in practice these are often just injective maps, not inclusions in the sense of set theory (for example, $\mathbb{Z} \to \mathbb{Z}[T]$). The category-theoretic definition might be abstract, but it is really useful in practice. Even abstract algebra texts such as the one by Atiyah-Macdonald use it without any explanation. So their definition of a subring is not just wrong because the closure under additive inverses is lacking, but also because this is not the definition which is used in the book.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.