# Why do Dummit & Foote require an ideal also be a subring?

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

• 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? Apr 27, 2021 at 6:54
• 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.
– user239203
Apr 27, 2021 at 6:57
• Thanks for your comments! It turns out D&F doesn't require rings to be unital. Apr 27, 2021 at 7:01
• See also wiki for a short summary of the debate ring-with-1 versus rng.
– user239203
Apr 27, 2021 at 7:09