The task is to construct a field of order 25.
works because it yields , right? It's clearly order 25, and is irreducible (it doesn't have a root in ) and thus is maximal so clearly is a field. What I'm wondering is just this: if there is more than one irreducible quadratic, does it make a difference which one we mod by? I don't think it does but I'd like confirmation. Thanks.
Thank you for the answer. I'd like to make sure I understand this. So . Modding by this ideal yields all polynomials with degree less than that of . But then why exactly doesn't it matter which polynomial (ideal) we mod by? Modding by a given polynomial makes all multiples of this polynomial "0," but then what about the other irreducible, say, quadratics? I know this isn't well-posed at all but hopefully someone will see what I'm trying to ask...
I guess I could also ask, what happens to elements in the other ideals generated by the other irreducible quadratics, that do not have as a factor? I'm just missing something (obviously).
It matters, but not up to isomorphism. It is an interesting exercise to prove that if f and g are irreducible polynomials of the same degree then . That is, to prove that there is only one field of order for a given p and n. (Actually, I think this is quite hard to do in general - so perhaps a better exercise would be to pick a prime, 5 say, and two irreducible polynomials, of degree 2 say, and find the isomorphism. It is tricky-ish, but interesting.)
the idea is, if you mod by an irreducible (monic) polynomial f(x) of degree n, then for any root a of f, a^n can be expressed as a
linear combination of 1,a,a^2,...,a^(n-1).
if our underlying field is F, we obtain a field of dimension n (as a vector space) over F, isomorphic to F^n.
if |F| is finite, then the extension has order |F|^n elements (|F| for each basis element).
we know that the powers of a up to a^(n-1) have to be linearly independent over F, because otherwise a would satisfy some other polynomial
in F[x] of degree < n, say g(x). writing f(x) = g(x)q(x) + r(x) (using the division algorithm) we have:
0 = f(a) = g(a)q(a) + r(a) = 0 + r(a) = r(a), that is g(x) divides f(x) in F[x], contradicting the irreducibility of f(x).
now, the fields we get by using two different polynomials aren't necessarily "the same", but they ARE isomorphic.
let's look at what we get when we have E = Z5(a), when a is a root of x^2 + x + 1, and when K = Z5(b), when b is a root of x^2 + 2.
what we are looking for is some s + ta in E that satisfies: (s+ta)^2 = 3 (-2 in Z5). remember in E, a^2 = 4a + 4.
so let's see what we get:
(s + ta)^2 = s^2 + 2st(a) + t^2(a^2) = s^2 + 2st(a) + t^2(4a + 4)
= (s^2 + 4t^2) + (2st + 4t^2)a = 3. for this to be true, 2st + 4t^2 = 0.
since 3 is not a square in Z5, t is non-zero, so we have 2s + 4t = 0.
thus 2s = t. we also have to have: s^2 + 4t^2 = 3, that is:
s^2 + 4(2s)^2 = s^2 + (4)(4)s^2 = s^2 + s^2 = 2s^2 = 3.
so s^2 = 4 --> s = 2 or 3. so either 2+4a or 3+a is a square root of 3 in E.
thus the desired isomorphism from K to E sends 1-->1 and b-->2+4a, or b-->3+a. see how that works?
Ok, let's back up.
we start with a field, F, and then we consider polynomials in x over F, that is, F[x]. F[x] is a ring, in fact it is an integral domain.
now we have the ideal of F[x] generated by f(x), I = <f(x)>. if f(x) is irreducible, than I is maximal, which means that F[x]/I is a field.
consider the element a = x + I in F[x]/I. then:
f(a) = f(x+I) = f(x) + I = I = 0 + I, that is, in F[x]/I, f(a) = 0 (I = 0+I is the 0-element of F[x]/I).
now, what does a typical element of F[x]/I look like? well, suppose deg(f) = n.
then every element of F[x]/I is of the form b0 + b1x + b2x^2 +....+ b(n-1)x^(n-1) + I =
(b0 + I) + b1(x + I)^2 +....+b(n-1)(x + I)^(n-1) = b0(a) + b1(a^2) +....+b(n-1)(a^(n-1)).
in other words, F[x]/I is isomorphic to F(a), where a is a root of f(x).
put another way: considering F[x]/<f(x)> is a way to get the root of f(x) we need, the coset x + <f(x)> is automatically a root of f.
the field we get doesn't "really" depend on what f is, it only depends on the degree of f, in the sense that we can
obtain an isomorphism of E = F[x]/<f(x)> with F[x]/<g(x)> = K, when deg(f) = deg(g) (and both are irreducible).
for example, one way to construct the complex field is to consider R[x]/<x^2 + 1>.
the coset x + <x^2 +1> can be considered as the complex number "i" (or -i), that is, a root to the polynomial x^2 + 1 = 0.
The typical way to show uniqueness of finite fields is to note that a field of size is a splitting field for the polynomial over its base field (because the multiplicative group of a finite field is cyclic of order , every element satisfies that polynomial). Since splitting fields are unique up to isomorphism, so are finite fields.
this is true, however, usually the proof of uniqueness of a splitting field, uses in an essential way the construction of F(a) where a = x+<f(x)> in F[x]/<f(x)>.
i mean, there is no getting around this (well, for extensions of Q, you can restrict your attention to subfields of C, and invoke the algebraic closure of C.
it is interesting that there is no algebraic proof of the so-called Fundamental Theorem of Algebra). but since we are talking about fields of finite characteristic,
results involving the rationals aren't much help.
this particular construction, taking a quotient of a polynomial ring by a principal prime ideal, seems to give people trouble.
no it doesn't have a root in F. it has a root in an extension of F. historically, what they used to do was "adjoin a root":
so you would create a field generated by F and the root of f(x), a, called F(a). we can define this as the smallest field containing F and a.
what does this field look like? well, suppose we take this detour:
consider the mapping φ:F[x]--->F(a) given by φ(h(x)) = h(a). this is clearly a ring homomorphism. so, we can ask, what is ker(φ)?
ker(φ) = {g(x) in F[x]: g(a) = 0}. clearly f(x) is in ker(φ). in fact, any g(x) in ker(φ) has to have degree ≥ deg(f) (if a satisfied some polynomial p(x) in F[x] of lesser degree, then writing f(x) = p(x)q(x) + r(x), we have:
0 = f(a) = p(a)q(a) + r(a) = 0 + r(a) = r(a), so a also satisfies r(x) where deg(r) < deg(p). we could then divide f(x) by r(x), eventually (after a finite number of steps) arriving at:
f(x) = m(x)s(x) + r for some polynomials m,s where m(a) = 0, and r is in F.
then 0 = f(a) = m(a)s(a) + r = 0 + r = r, which is a contradiction because f is irreducible).
so every element of ker(φ) has degree ≥ deg(f). again, you can use the division algorithm to show that every element g(x) in ker(φ) has f(x) as a factor, so
ker(φ) = {h(x)f(x) : h(x) in F[x]} = <f(x)>.
by the first isomorphism theorem (for rings), φ(F[x]) is isomorphic to F[x]/<f(x)>.
for any field element s (i.e, a constant polynomial), φ(s) = s. also, φ(x) = a. now since f(x) is irreducible, <f(x)> is a maximal ideal, so the only ideals of φ(F[x]) are {0} and φ([F(x)]), so φ(F[x]) is a field containing F and a, since φ is onto its image, it must be the case that φ(F[x]) IS F(a).
That does clear it up. I haven't learned aboute extension fields yet, so I wasn't thinking in that way. I will continue thinking about this, but I don't want to beat it to death. Need to keep moving forward! Thank you for all the help.