Talk:Ring ideal

I took this out for now:

Alternatively, all of the requirements may be replaced by the following: any finite R-linear combination of elements of I belongs to I

I don't particularly like this, because it hides too many things. For instance, we would have to have the understanding that linear combinations are two-sided linear combinations, whereas in the typical vector space and module setting they are only left-sided combinations (and two-sided linear combinations don't make sense). Furthermore, we would have to spend a paragraph explaining that finite combinations include combinations of zero elements which are defined to be the zero element of the ring that the zero elements were chosen from. In summary, I think this alternative definition looks cuter than it is. AxelBoldt, Sunday, June 9, 2002

It's not meant to be cute; it's meant to show that there is a single broad class of operations — the linear combinations — that ideals are closed under. I know that thinking in these terms makes ideals (and, more generally, modules) clearer to me, but I agree that it's a less elementary point of view. I would be happy to move the comment to the paragraph that mentions the relationship between ideals and submodules. Since linear combinations are inherently central to module theory (that is, linear algebra), this is an appropriate place; additionally, this comes after we've discussed the various flavours of ideals, so that a quick parenthetical "(where the linear combinations are on the left, on the right, or two-sided, accordingly as the ideal)" will take care of that. In any case, I think that it's worth mentioning somewhere, even if way at the bottom; the same thing on the pages Submodule[?] and Vector_subspace (or Module and Linear_algebra/Subspace, which is where those topics are hiding out now). As for the zero linear combination, that can be mentioned on the page Linear_combination (once it exists — I was shocked to see a red link in my Preview!). After all, if anybody is confused about how closure under linear combinations could yield the zero element, then that's what they'd look up, right? — Toby Bartels, Tuesday, June 11, 2002

PS: Hey, no more red link! Needs work, however. — Toby

I'm happy now if you are. The next step is to work on Linear_combination ^_^. — Toby Bartels, Tuesday, June 11, 2002