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From Finite Integral Domains to Finite Fields
- revskill 1y agoSo what is the point of being a field ?
- thehumanmeat 1y agoYou get "division".
- Koshkin 1y agoFields are easier to deal with.
- markisus 1y agoIt’s an abstraction that helps mathematicians study interesting phenomena. I believe the famous squaring the circle problem was resolved using the language of fields.
- btilly 1y agoThat we can't square the circle comes from pi being transcendental. The result that you're thinking of is Galois' proof that there is no algebraic formula forroots of 5th degree polynomials.
- mathgradthrow 1y ago"transcendental" is field language
- btilly 1y agoI've always thought of "transcendental" as number theory language, though I can see how someone could argue that it is field language. But the Galois group of a field extension definitely is field language.
- mathgradthrow 1y agoa field extension is the thing which is transcendental or not.
- cka 1y agoYeah, and constructability is usually handled by proving that a length is constructable if it lives in an iterated quadratic extension of the rationals. Pi does not lie in such an extension, so is not a constructable length (and neither is its square root).
- inglor_cz 1y agoOver fields, polynomials mostly behave as expected, and systems of linear equations are solved very similarly to R. Basically, you can adapt quite a lot of real and complex algorithms to other fields, including matrix operations. Once you leave fields and then even integral domains, things get weird. For example, the quadratic equation x^2 = 1 has four roots in Z_8.
- vouaobrasil 1y agoThe high level answer is that every module over a field is free. That is, if F is a field and M is an F-module then M is isomorphic to a direct sum of F, which may be a finite or infinite direct sum.
- getnormality 1y agoSee also Wedderburn's little theorem, which shows that any finite division ring is commutative and therefore a field. This is a pretty amazing result because rings were created partly to study algebra in a non-commutative setting, and many of the most important rings, such as n x n real matrices with n > 1, are non-commutative. The quaternions in particular are a non-commutative division algebra, not subject to the theorem because infinite. The proof of Wedderburn's little theorem is relatively simple by the standards of professional math, but it's beyond me to even imagine ever coming up with it.
- curtisszmania 1y ago[dead]
- clintonc 1y agoYou can get that every integral domain is a field with fewer words by using a higher powered set theory result -- injections on finite sets are also surjections. The cancellation property says multiplication by any element is an injection, so it is also a surjection, i.e., 1 is in the range, so that gives you the multiplicative inverse.
- csense 1y ago> injections on finite sets are also surjections Not necessarily [1]. I think you're missing an assumption there. [1] https://en.wikipedia.org/wiki/Injective_function#/media/File:Injection.svg https://en.wikipedia.org/wiki/Injective_function#/media/File...
- MaxRegret 1y agoIn this case, multiplication by any nonzero fixed element of the ring is an injection from the ring to itself. Any injection from a finite set to itself is indeed a surjection (and so also a bijection).
- susam 1y agoThe intended point, I believe, is the fact that any injective function from a finite set to itself is also surjective.
- vouaobrasil 1y agoThe correct statement is that an injection from a finite set to itself is a surjection. The converse is true, too. A surjection from a finite set to itself is an injection.
- Tainnor 1y ago> Let F be a field, and let a,b∈F such that ab=0. There are two cases to consider: a=0 and a≠0. If a=0, then indeed ab=0 by Proposition 1. This part is a bit weird. If a=0, then we are already done, there's no need to prove ab=0 (which was already the assumption). The other case can also be proved in a shorter way by just multiplying both sides of ab=0 with a^(-1) from the left.