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1. All coefficients are taken modulo 3, so x^2 + x - 1 is equal to x^2 + x + 2. 2. Yes. Again, 4x+4 = x+1 over Z_3.

@@katherinekelm7439 Right! Thank you! I missed the Z_3, I'm unused to finishing with a modulo...

@@katherinekelm7439 So what would happen if we had a polynomial like 3x+3 over Z_3, would it all turn into 0 since 3 mod 3 = 0 or x+1 ?

@@pedromendes6846 Yes, the coefficients are actually congruence classes in Z_3, so 3x+3 is actually [3]x+[3] = [0]x+[0].

Yes. Every linear combination of a and b is a multiple of gcd(a,b), and every multiple of gcd(a,b) can be written as a linear combination of a and b.

​@@katherinekelm7439 I am going to use this set result to produce a corollary. Note that i will use (a,b) in place of gcd(a,b) for readability. So to recap , given integers a,b not both zero and defining the sets A and B , A = { u | u = ax + by, x and y are integers } , B = { v | v = k* (a,b) , k is an integer }, we can prove A = B , using two set inclusions ( i will not attempt that but it can be shown). We immediately get the following corollary, for integers a,b not both zero, "Given integers x,y such that ax + by = 1, it follows (a,b) = 1" Proof: By the set equality, we get ax + by = k * (a,b) for some integer k . Since we are given ax + by = 1 , by substitution 1 = k * (a,b) . but (a,b) >= 1 since a,b are not both zero, whence it follows that k must be 1. (I think this makes sense, since k is an integer and the only way to factor 1 into a pair of integers is 1*1 or (-1)*(-1) ). Thus (a,b) = 1. Maybe this is not the customary presentation of this number theory result, or the most efficient, but i do like the set theory flavor , and the logic is transparent. Sorry if this seems pedantic, it feels more amenable to numerical exploration ; i can open a maple document and experiment with such results, comparing sets. As some mathematician once said, all of math can be reduced to sets.

Quite right. Good catch!

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@@katherinekelm7439 Thank you

It depends on how one defines Z_n. In most books Z_n is defined as the set of all congruence classes (mod n). So the class [k] in Z_n is exactly the coset k + nZ in Z/(nZ).

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If you use the Euclidean Algorithm to divide x³+6x+1 by x²+3, the last non-zero remainder is 3x+1. Factoring out 3, we get 3(x+5), so the gcd is x+5. You can check that x³+6x+1 = (x+5)(x² + 2x + 3) and x²+3 = (x+5)(x+2).

Thanks

The 'x' is considered "indeterminate", meaning it is just a place-holder and not an element of any field. F[x], then, is just a collection of objects called "polynomials" that just happens to have a ring structure when we add and multiply the polynomials in the usual way. There are some results that allow us to "plug in" field elements for x and get something meaningful -- for example, we can talk of a "root" of a polynomial f(x) which is an element c of F (or of an extension field of F) such that f(c) = 0 -- but the basic ring structure of F[x] is independent of such results. Hope that helps!

Ah I see, thanks for the detailed response! So it seems that these polynomials aren't really functions, huh. What might be the rigorous way to construct a function out of a given polynomial? What are possible options for what the domain of the function could be? (since there needs to be a notion of "multiplying x by a coefficient, or taking powers of it")

@@aditya_a If F is a field then you can always create a function f: F → F defined by a → f(a), where f(x) is a polynomial in F[x] -- in fact you could use f : K → K where K is an extension field of F, i.e. a field where it makes sense to add and multiply elements of F. For example, consider f(x) = x^2 - 2 with coefficients considered to be rational numbers. Let f : R → R (R = the set of real numbers) be defined by a → a^2 - 2 for each a in R. Since every rational number is also a real number, this is a well-defined function.