Ideals and polynomial rings

Either you really don't understand your own question or you've got to be kidding: this is way too long and complex (and interesting and beautiful) a theme to treat it here. You better go and grab a book (try "Algebra" by Dummit & Foote, or the book by Hungerford, or the one by Fraleigh) and read about this there.

DonAntonio

 

Here you go:
http://en.wikipedia.org/wiki/Ideal_(ring_theory)
http://en.wikipedia.org/wiki/Polynomial_ring

 

A basis of an ideal is a set of polynomials that "generate" the ideal; that is, if f1, .., fs is a basis for I, then each f in I can be written as h1f1 + h2f2 + ... + hsfs. A basis "spans" an ideal in a similar sense to how a linear basis spans a vector space.

Every ideal of a polynomial ring has a finite basis; this is called the Hilbert Basis Theorem.

Any ideal of polynomials in one variable can be generated by a single element. The ideal generated by f, a polynomial in x, is the set of polynomials which are divisible by f. Therefore, we can check for membership in the ideal just be computing that single polynomial generator, and then dividing (polynomial long division, which you may recall from high school, and writes g = qf+r, to divide g by our generator f). The remainder r is zero if and only if g is in the ideal generated by f.

In larger numbers of variables, this fails. There may be more than one generator; this means there are several different ways we can write "f = h1f1 + h2f2 + ... + hsfs + r," with several different remainders r. (There are algorithms for "dividing by multiple polynomials at the same time," or "dividing by a basis.") If we find one of these remainders, and it is zero, then we definitely have membership; if we find one and it is not zero, however, it doesn't mean that there is not a different way where the remainder is zero. We might hope for an alternative.

Groebner bases are a special basis, which every polynomial ideal has, with the extremely useful property that dividing by a Groebner basis gives remainder zero if and only if the dividend is in the ideal. This makes them very useful objects in commutative algebra and algebraic geometry.

 

....

 

Restrict my discussion to fields then. I'm in the middle of studying commutative algebra, so I'm definitely no expert. Thank you for bringing this up.

As for ideals and vector subspaces being similar, I would still argue they are. They are subobjects of an algebraic object with a given collection of operations. In each case they are closed under a particular subset of those operations, and the basis uniquely generates that subobject under those operations. (It's worth note that in each case, the subobject is comprised not just of operations and elements, but operations and two distinct types of elements.)

In the ideal case, the structure whose elements they can be "multiplied" by contains the ideal, and the vector space case, the elements by which we can "multiply" (the field itself) is in a different sense contained in the space.

Also, linear independence is not required in the ideal case. I would actually not consider this a difference; algebraic independence is required (for a minimal basis), which is a bit more parallel, seeing as the ideal case takes algebraic, not linear, combinations.

Those are two considerable differences, but I certainly wasn't trying to claim that there is a correspondence between the two. I only meant that, if we're not being too pedantic about it, it can be a helpful parallel for someone familiar with linear algebra, but not with ideals.

 

Buchberger himself has written things that give simple examples. As I recall, his thesis did. On the web, I found this talk: http://www.google.com/url?sa=t&rct=...7b2jCg&usg=AFQjCNFkMMyThqwYMzBCs4FPSt1qZofMEw

glebvog, Why don't you explain one of those examples to us! I need a review.

(If you're researching this topic with a view to solving some simultaneous polynomial equations, you should also look at "Wu's elimination method".)