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On page 242 of Abstract Algebra, 3rd Ed., by Dummit & Foote, they write:

Abstract Algebra, 3rd Ed., Dummit & Foote

However, in my lecture notes, there is a line that says

Definition. (Ideal.) Let $R$ be a ring. A subgroup $I$ of the underlying abelian group $R$ is called an ideal if $rx \in I$ for all $r \in R, x \in I$.

Note that $I$ may not contain [the multiplicative identity], so it may not be a subring. In fact, if $1 \in I$, then $I$ must be the whole ring!

Surely this is a typo in Dummit and Foote, right? Like my lecture notes suggest, if $1 \in I$ then $I = R$ since $r \in I$ for all $r \in R$.

Wikipedia seems to agree that an ideal need not be a subring.

EDIT: I realised Dummit & Foote doesn't require a subring to include the multiplicative identity. This answers my original question, but begs a new one: why would D&F define a subring to not necessarily be itself a ring?

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    $\begingroup$ I don't know the details of these two books - but is it possible one of them considers rings to be unital and one doesn't? $\endgroup$
    – preferred_anon
    Apr 27, 2021 at 6:54
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    $\begingroup$ You should check if the the book defines "ring" as "ring with $1$" or not. If the former, then subrings should also contain $1$ (you might want to check this definition as well) and it should be a typo. If the latter, then the current version is the intended statement by the authors: you'll just have to accept that the textbook and your lecturer are using different notations and that this might come up upon further reading. $\endgroup$
    – user239203
    Apr 27, 2021 at 6:57
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    $\begingroup$ Thanks for your comments! It turns out D&F doesn't require rings to be unital. $\endgroup$
    – Jeremy Lindsay
    Apr 27, 2021 at 7:01
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    $\begingroup$ See also wiki for a short summary of the debate ring-with-1 versus rng. $\endgroup$
    – user239203
    Apr 27, 2021 at 7:09

1 Answer 1

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As the comments on the original post suggested, the book by Dummit & Foote does not require rings to be unital (contain a multiplicative identity). Naturally — and to answer the question in the edit — subrings need not be unital rings either, according to Dummit and Foote.

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    $\begingroup$ What you've written is perfectly correct, but just another note: that's not to say that the ideal doesn't have an identity. It's just that proper ideals will certainly not share the same identity as the containing ring. $\endgroup$
    – rschwieb
    Apr 27, 2021 at 13:20

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