Wikipedia:Reference desk/Archives/Mathematics/2016 May 16

= May 16 =

Rational functions that can't be evaluated anywhere
I was thinking about one of my responses to DHeyward in the "Infinity/infinity" thread above. The point was that, while the identity x2/x=x works only for x&ne;0, when you're evaluating at real (or complex) numbers, nevertheless it is perfectly valid to write x2/x=x, without restriction, in the field of rational functions over the reals. Here, we aren't evaluating at points; rather, we're dividing one polynomial by another.

So that got me to thinking about what happens in other rational-function fields. For example, if we write Z5 for the field of integers mod 5, then (someone will correct me if I'm wrong), x5-x is a perfectly fine polynomial, distinct from the zero polynomial, even though it equals zero when you evaluate it at any element of Z5.

Continuing along this line, 1/(x5-x) is a perfectly fine rational function over Z5, even though it does not have a value at any point.

Is this the standard exposition of rational functions over finite fields? Or is it generally useful to identify polynomials that are everywhere equal? Is there a name for the structure where you do make that identification? --Trovatore (talk) 20:35, 16 May 2016 (UTC)
 * On a related subject, I was once asked to construct the finite field with four elements. I was asked to (as normal) use polynomials over the finite field with two elements, but got stuck when I realized that $$x^2 + x + 1 = 1$$ for both 0 and 1. The explanation I was given that this polynomial is not the same as the constant polynomial 1, even though they are everywhere equal.--Jasper Deng (talk) 20:59, 16 May 2016 (UTC)
 * I guess if you identify polynomials that are equal everywhere, you don't have an integral domain anymore, so you can't form the field of quotients. In my example, x5&minus;x factors into x&middot;(x&minus;1)&middot;(x&minus;2)&middot;(x&minus;3)&middot;(x&minus;4).  So if you were to identify it with zero, you would have for example that neither x&middot;(x&minus;1) nor (x&minus;2)&middot;(x&minus;3)&middot;(x&minus;4) is zero, but their product is zero.  That would be inconvenient.
 * If you identify polynomials that are everywhere equal, you end up with the set of all functions from the finite field to itself (over finite fields there are more polynomials than functions, while over infinite fields there are more functions than polynomials). —Kusma (t·c) 10:51, 17 May 2016 (UTC)


 * I'm thinking that the term "field of rational functions" is really a misnomer. You should think of x not as a variable but as a transcendental constant. This seems a bit more natural when the base field is Q; if t is a transcendental number then the field you get by adjoining t to Q is isomorphic to Q(x), where the isomorphism is determined by x↔t. In other words Q(π), Q(e), Q(ln2) are all the same algebraically, so you might as well replace the π, e, ln2 with an indeterminate x, and you shouldn't think of 1/(x5-x) as function any more than you think of 1/(e5-e) as a function. For Z5, think of it as being embedded in a field with transcendental elements. Take one of these elements and call it x, the choice is arbitrary but once made x is a constant. The smallest subfield containing x is then isomorphic to the field of rational functions Z5(x). --RDBury (talk) 11:58, 17 May 2016 (UTC)


 * This use of "function" (for something that isn't one) pops up in a few other places, notably "generating function." --JBL (talk) 18:04, 17 May 2016 (UTC)
 * Well, there is a (partial) function naturally associated with it, though. You can reduce the fraction to lowest terms, then evaluate the numerator and denominator at each point of the underlying field and divide.  In the case I gave, that turns out to be nowhere defined, but in general it still seems to be a natural interpretation. --Trovatore (talk) 18:15, 17 May 2016 (UTC)