M BUZZ CRAZE NEWS
// updates

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

By Gabriel Cooper
$\begingroup$

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?

$\endgroup$ 4

1 Answer

$\begingroup$

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.

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

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