[HN Gopher] From Finite Integral Domains to Finite Fields
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From Finite Integral Domains to Finite Fields
Author : susam
Score : 53 points
Date : 2025-05-26 13:43 UTC (3 days ago)
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| revskill wrote:
| So what is the point of being a field ?
| thehumanmeat wrote:
| You get "division".
| Koshkin wrote:
| Fields are easier to deal with.
| markisus wrote:
| It'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 wrote:
| That 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 wrote:
| "transcendental" is field language
| btilly wrote:
| I'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.
| cka wrote:
| Yeah, 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 wrote:
| Over 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 wrote:
| The 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 wrote:
| See 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.
| clintonc wrote:
| You 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 wrote:
| > 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...
| MaxRegret wrote:
| In 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 wrote:
| The intended point, I believe, is the fact that any injective
| function from a finite set to itself is also surjective.
| vouaobrasil wrote:
| The 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.
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