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So one way to define a group presentation is to say, well, let's generate the free group with some number of generators, and then quotient by saying certain elements (relators) cancel just as $aa^{-1}$ does (and other things, you have to take the normal subgroup generated by the elements, but the basic idea works, I think). That is, we have for some words $R_1, R_2, \ldots$ the relations $R_1 = R_2 = \ldots = e$.

A simple example is the cyclic group of order n: one presentation is $\langle g \mid g^n \rangle$.

My question is: can and do we do this for rings? I'm imagining you would do something like take the "free ring" consisting of all sums of products of generators, and perhaps instead of having all the relators equal to the trivial group they would equal the zero ideal, because then if you wanted some word $R$ to equal the multiplicative identity you would just say $R - 1 = 0$. (In particular, the comments on the vaguely related question that got me thinking about this is here: could you formulate a rigorous proof by "dividing by the relations"?)

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You sure can. Both for algebras and rings we have this notion. For rings we can see it as a polynomial ring: en.wikipedia.org/wiki/Polynomial_ring#The_polynomial_ring –  BBischof Dec 2 '10 at 21:28
    
Oh, that was obvious. Whoops. –  Harry Stern Dec 2 '10 at 21:53
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For noncommutative setups, the tensor algebra is used rather than the polynomial algebra. –  user1119 Dec 2 '10 at 22:48
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An interesting exercise for you: Construct the coproduct for two rings. First try in the category of commutative rings, and then in the category of possibly noncommutative rings. –  user1119 Dec 2 '10 at 23:34
    
@George S. yes :) –  BBischof Dec 3 '10 at 6:09

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up vote 13 down vote accepted

You certainly can: the constructions of the tensor algebra, symmetric algebra, and exterior algebra can all be performed using generators and relations (though are perhaps usually thought of conceptually via a universal property). The operation of localization is also an example of this: one formally adjoins "denominators" satisfying the requisite relations (localization of $A$ at a multiplicative set $S$ can be thought of adjoining variables $x_s, s \in S$ with the relations $s x_s = 1$).

The notion of a ring having a "finite presentation" over another ring $R$ (i.e. be the quotient of a polynomial ring $R$ by a finitely generated ideal--that is, by finitely many generators and relations) is of importance in algebraic geometry, as many properties proved in, say, the finite type case over noetherian schemes extend here (e.g. I believe Chevalley's theorem of the constructibility of the image of a constructible set under a morphism of schemes is true under finite presentation hypotheses). I am not sufficiently qualified to say much more on this topic, but you might find BCnrd's answer here enlightening.

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I am not that advanced yet (I'm a junior in high school), but I have at least seen some of the names of those constructions in books I'm reading, so I guess I'll get there eventually. Thank you! –  Harry Stern Dec 4 '10 at 1:45

You can consider presentations for any algebraic system given by special constants, operations and equations. For example, groups are given by the special constant 1 (for the multiplicative identity), the unary operation of inverse, x-1, and the binary operation of multiplication, $xy$, the plus the equations that these must satisfy. In that case, you can specify every object with such operations in terms of a set of generators, and the relations that the generators must satisfy. The study of these kinds of constructions is known as universal algebra.

For the specific case of commutative rings over a field, there is an explicit algorithm for working with presentations, known as the Groebner basis algorithm. It's widely implemented in symbolic algebra packages. (In the case of commutative rings, you're basically just working with systems of polynomials.)

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