Abstract Algebra Leftovers

Normal Subgroups

Having discussed subgroups and kernels of groups, one can make some statements about the order of subgroups in relation to that of the parent group. Of course generally, the order of the subgroup is never larger than that of the parent group, but we can segment our groups through congruence to arrive at a very important Theorem.

We remember the definition of congruence classes as the subset of a set which are equal to one another modulo some element of the group. Such an element can be taken as a generator of a group, so the definition for subgroups K of groups G as modulo becomes that with b* the inverse of b: ∀ a, b ∈ G : ab* ∈ K <-> a is congruent to b. This congruence relation retains reflexivity, symmetry and transitivity that it already had in non-group notation, along with all other properties. We keep those in mind as we go on.

More analogues show up in the definitions of (right) cosets Ka. We remind ourselves that cosets are the compositions of all elements of the subgroup K with the element a ∈ G, i.e.

Ka = { ka : k ∈ K }

By group axioms and cancellation, Ka then forms a subgroup of G (the union of all Ka are also a subset of G). Since the neutral element of operations is always included in any subgroup of G, then each element g ∈ G is at least in the right coset Kg, meaning that the union of right cosets are exactly G. This is of course also true for the left cosets. We can also map every element of K to a unique element of Ka by associating k -> ka ∀ k ∈ K, so there exists a bijection between K and all of its right cosets. They have the "same number" of elements. The number of cosets of a subgroup K in G is called the index of K in G. We write [K:G].

Since we know that the union of all right cosets of some subgroup K is equal to G, we can then easily see that |G| = |K| [G:K]. This is Lagrange's Theorem and it will make talking about groups and the orders of their elements really convenient. Take any element g of a finite group G. The order n of a will generate a cyclic subgroup <a> with order n. Since <a> is a subgroup of G, Lagrange's Theorem applies and n will divide |G|. For groups of prime order, this also means that for every subgroup K, either |K| = 1 or [G:K] = 1. For any cyclic subgroups of non-identity elements in K, one should hope that |K| > 1, so [G:K] = 1, meaning that K is all of G. This makes every group of prime order isomorphic to Zp (the group of real numbers up to p). For groups of non-prime order, these might be divisible into products of groups of prime order. The details of this will come up later.

Having a good definition of right cosets, we can define some subgroups with special traits. Chiefly, we can picture in abelian groups that all right cosets of some subgroup are equal to their left cosets. Such subgroups are said to be normal subgroups. There is a number of equivalent conditions to identify normal subgroups that all come down to canceling of the expression aN = Na for some subgroups N and ∀ a ∈ G, and aren't otherwise interesting, so I won't go into detail here. You can easily find these on cursory google searches, if you're curious. I would advise tracking down a definition that seems useful to you and stick to that one.

Using the normal subgroup, we define G/N, that is the set of all right cosets of N in G. Because of the bidirectionality of the transitivity of congruence in normal cosets, we can define an operation so that G/N comes out as a group: (Na)(Nb) = Nab

Now let's assume that G is finite. |G/N| is the number of distinct right cosets of N, which we would ordinarily identify as the index of N in G [G:N]. Lagrange's Theorem says that |G| = [G:N] |N|, so |G/N| = [G:N] = |G| / |N|. Because of the bi-directional construction of the normal subgroup, commutativity is inherited to N from G. Commutativity in G/N can also be characterized through aba*b* ∈ N, where * denotes the inversion of elements. This follows from Nab = NaNb = NbNa = Nba and Nab = Nba, meaning (ab)(ba)* ∈ N, i.e. ab and ba are congruent. Subgroups K of G that contain N inherit the subgroup relation to K/N.

A special case, where G/Z(G) is cyclic then automatically makes G abelian, since Z(G) is normal by definition and we can calculate that aba*b* is in Z(G) directly.

We've already had allusions to the role of isomorphisms and homomorphisms in these special subgroups, so let's make these relationships explicit. Let's first define the kernel of the group homomorphism f: G -> H. ker(f) = { a ∈ G: f(a) = e }. It constructs a normal subgroup in G. This is mainly because you can do very literally anything with the identity element and ker(f) only projects onto the identity element. The group axioms follow trivially in H, and the elements in ker(f) need to retain the structure of the groups. Because of this, only the identity element of G maps onto the identity element of H if and only if f is injective. Not only that, but group homomorphisms with the kernel K have f(a) = f(b) if and only if Ka = Kb. We are almost at the point where we can identify abstractly which groups and subgroups are isomorphic to one another. To do this, we should define a map p(a) = Na. It maps from G -> G/N and is a surjective homomorphism with the kernel N. The product follows directly from the product of right cosets and the solution of the equation p(a) = Ne.

A surjective homomorphism f: G -> H with ker(f) = K lets us define q: G/K -> H, q(Ka) = f(a). Choose a, b so that Ka = Kb, then f(a) = f(b), i.e. q(Ka) = q(Kb). q then is well defined. Surjectivity of q follows from the surjectivity of f, and injectivity follows from construction. q(KaKb) = q(Kab) = f(ab) = f(a)f(b) = q(Ka)q(Kb), so q qualifies as a homomorphism. Then q: G/K -> H is an isomorphism.

Assume K and N are normal subgroups of G and N is a subgroup of K. K/N inherits the traits of a normal subgroup in G/N and (G/N)/(K/N) is isomorphic to G/K. Choose a and c in G/N so that they have congruence, then ac* ∈ N. f: G/N -> G/K, f(Na) = Ka is well defined as already shown, so bijectivity and homomorphism properties remain. The latter is easily shown explicitly as it went above. Surjectivity is trivial. Injectivity follows from f(Na) = Ke -> Ka = Ke for Na ∈ ker(f). These are the first and third isomorphism theorems respectively.

Together they imply that normal subgroups of normal subgroups are automatically normal.

Group Theory

This is where the real meat of abstract algebra begins. We have the objects and vocabulary to classify finite groups. We'll now work with groups as objects rather than their elements. First, that means that we want to be able to have groups interact beyond the means of isomorphisms, which are really more of a relation, rather than an operation. Considering the non-trivial structure of groups, it should make sense to define an outer product (or rather, Cartesian product) of two groups G × H. We define this analogously to the Cartesian product of objects like linear spaces, defining it as a set of all possible permutations of all the elements in an n-tuple. This means that groups that are included in the Cartesian product are not subgroups of that product just by inclusion. There are however some cases, in which a proeduct of normal subgroups comes out isomorphic to their parent group.

We take two normal subgroups N, M of G, whose intersection is exactly the neutral element. The elements of those subgroups are commutative respectively. We only check this for the case that the elements are from different subgroups, because other other cases follow directly from the definition of normal subgroups. Let a ∈ M, b ∈ N and consider a*b*ab. By normality, b*ab ∈ M. By closure, a*b*ab = a*(b*ab) ∈ M. Analogously, (a*b*a)b ∈ N. It then is part of the intersection of N and M and thus exactly the neutral element.

Now let's take an arbitrary number of normal subgroups of G such that every element g ∈ G has a unique product form consisting of exactly one element of each normal subgroup. We then define a map to collect these terms, which naturally maps onto G. This map is by assumption surjective and by uniqueness injective. Such a construction is only possible for subgroups whose pairwise intersection is exactly the neutral element (we call such groups mutually disjoint), as there would be at least two representations of products for at least one element of G. In sum, this makes the defined map an isomorphism.

An example of direct addition ⊕ working as an operation on groups instead of direct multiplication are the finite Abelian groups. We apply the common substitutions here, and with the same logic the statements from the direct products carry over without change. We can define an abelian group of some prime p so that G(p) = { g ∈ G : |g| = p^n, n ≥ 0 }. Under addition, this set fulfills the group axioms.

Let's take a look at some element of finite order in g ∈ G (all elements in finite group have finite order, remember?) and the set of primes that divide g. g is then equal to the sum of some elements of the finite abelian Groups of those primes and every finite abelian group is equal to the direct sum of finite abelian groups of distinct positive primes that divide the order of their sum. So far all this is familiar by intuitively translating the operator onto the same space. In the following paragraphs I will refer to these properties as "basic properties" of finite abelian groups.

By these properties, we can tell that any finite abelian group G is a direct sum of its sub-p-groups G(p) with each prime that divides |G|. Any finite abelian p-group can be decomposed into a direct sum of cyclic groups, as any group of order p is cyclic and isomorphic to Z/p. We take an element g of maximal order in G, then there exists a subgroup K with G = <g> ⊕ K. This is the fundamental theorem of finite abelian groups.

Through the uniqueness of the decomposition into primes, we can make the general claim that for (n,m) = 1, the group Z/nm is isomorphic to Z/n ⊕ Z/m. Of course this extends to products of Z/p where p are powers of distinct primes. This makes every finite abelian group a direct sum of cyclic groups of orders that are pairwise coprime. These integer orders are called invariant factors of G. If these happen to be prime powers, they are called elementary divisors for their property of not being able to be reduced to smaller cyclic groups.

This representation is interesting to us for its ability to highlight isomorphisms. All groups that share the exact set of elementary divisors are isomorphic. Of course the direct sum commutes, so one direction of this proof is trivial by reordering. Assume G and H are isomorphic (by some function f: G -> H), then |a| = |f(a)| ∀ a ∈ G. This includes primes, and so f(G(p)) = H(p), so G(p) ≅ H(p). This then reduces our problem to showing that p-Groups have the same elementary divisors if they are isomorphic. Obviously this same property extends to groups with the same invariant factors.

Abandoning commutativity in finite groups will generalize the properties we've seen proven for the abelian case. For this, we use another kind of special subgroup, called the Sylow-Subgroups, which are characterized by the Sylow Theorems. Initially, we take some finite group G and a prime p where n is the largest power of p that divides |G|. Any subgroup of G with the order p^n is a Slow p-subgroup. We list the three Sylow-Theorems with minimal proof:

Let G be a finite group and p a prime so that p^k divides |G|. G then has a subgroup of order p^k

For all Sylow p-subgroups P, K of G there is an g ∈ G : P = x*Kx

The number of Sylow p-subgroups divides |G| and has the form 1 + pk, where k is some nonnegative integer

From the first Sylow Theorem it follows directly that finite groups whose orders are divisible by some prime p contain an element of order p, as there is a (cyclic) subgroup with order p. This is Cauchy's Theorem. This means that G = PQ and as P≅Z/p, Q≅Z/q, this concludes the proof.

We shall at least try and sketch out a proof for the Sylow theorems that we have so far used axiomatically. For this, conjugacy is a central tool. Conjugacy takes the familiar construction of Normal groups and applies it to elements. In a group, two elements a, b in G are conjugate, if there is an element x in G, so that a = x*bx. By canceling, this is of course symmetric and reflexive, and transitivity is given through (xy)* = y*x*. Conjugacy on groups then are equivalence relations and thus induce conjugacy classes. They are mutually disjoint with other conjugacy classes. We do this for several other concepts, such as the centralizer C(x), takes elements of the group that commute with x and construct a subgroup out of them. We will see in the future that primes that divide the order of finite abelian groups also provide the order to at least one element of the group. There is a longer, proof "by foot" so to speak, but it's not very interesting in the applications of previous theorems and mostly busywork. These concepts can be extended to subgroups, yielding conjugate subgroups and normalizers. We note, that conjugacy classes of G and |Z(G)| are disjoint. And we can write |G| as the sum of |Z(G)| and the order of all conjugacy classes.

A few further applications of the Sylow Theorems, now that we have a slightly expanded vocabulary. Assume G is a group of order p^n with some prime p and n > 0. By Lagrange's Theorem, |Z(G)| is a power of p less or equal to 0. |Z(G)| = |G| - |C1| - |C2| - ... where Ci are the conjugacy classes of G and their orders are larger than 1 dividing |G|. Because |G| is a power of a prime, its divisors larger than 1 are also powers of that prime. p then divides |Ci|. p also then divides |Z(G)| and it then can't be smaller than p, i.e. contains more than one element. However, we can also say that |Z(G)| is a normal subgroup and then Z(G) is either not equal to G, or G is abelian. This disqualifies G from being simple.

This tells us that groups of order p^2 are abelian, seeing as the order of Z(G) is either p or p^2. In the latter case G = Z(G), which makes G abelian. Otherwise |G/Z(G)| = |G|/|Z(G)| = p, which makes G a p-group and hence cyclic. This too would make G abelian. G then is isomorphic to Z/p^2 or Z/p × Z/p by the fundamental theorem of finite abelian groups.

For two distinct primes p and q with q≢1 (mod p) and p^2≢1 (mod q) and G a group of order (p^2)q, then by the third Sylow Theorem, the number of Sylow p-subgroups is congruent to 1 (mod p) and p divides |G|. 1 is the only possible option at this point, making the Sylow p-subgroup P unique, which is normal. For the unique Sylow q-subgroup Q, the argument works analogously. The order of the intersection of P and Q has an order dividing |P| and |Q|, which are coprime, so it's exactly <e>. P × Q = G, as we've seen in a previous proof. P is isomorphic to Z/p^2 or Z/p × Z/p and Q is isomorphic to Z/q, making G isomorphic to either Z/p × Z/p × Z/q or Z/p^2 × Z/q. This also implies that G is not a simple group.

Arithmetic in Integral Domains

We've not actually done this yet, and it's not quite trivial, even if the beginnings are easier to get into than last time when we were doing the group theory stuff. I will presuppose the basic definitions of and around integral domains and in general, we will assume R to be one such integral domain.

First, integral domains are interesting insofar as we can make a very helpful remark about units in R. Take u ∈ R to be a unit with an inverse v. For any r ∈ R: u(vr) = (uv)r = 1r = r, so units in integral domains divide every element. Let a, b ∈ R so that a is an associate of b. This means that a = ub -> va = vub = 1b = b. Since inverses of units are also units, then b also is an associate of a. If we assume that a is non-zero, and a = ub for all its associates b, it's trivial to see that a is divisible by all of its associates. We keep these properties in mind for all the propositions in this chapter.

A more specific domain is the Euclidean domain. An integral domain R is Euclidean, there exists some function f: R -> Z/+ so that ∀ a, b ∈ R non-zero, then f(a) ≤ f(ab), and if b ≠ 0 -> there exist some q, r ∈ R: a = bq + r and either r = 0 or f(r) < f(b). Such a function need not be unique. For units in Euclidean domains, we can make two statements for equivalency. First, f(u) = f(1). This is due to f(u) ≤ f(uv) = f(1), f(u) ≤ f(ub) for the inverse v of u, but also f(1) ≤ f(1u) = f(u). Second, f(c) = f(uc) for some nonzero c ∈ R, since f(c) ≤ f(uc) by definition and f(uc) ≤ f(vuc) = f(c). We also want to slightly change the definition of the greatest common divisors (gcd) in the Euclidean domain. It is of course a common divisor, but instead of identifying a greatest one, we extrapolate that for c | a (-> c = an) and c | b (-> c = bm) with (a, b) = d, then f(c) ≤ f(d) because d is a multiple of both a and b and c divides both. Similarly, the associates of every greatest common divisors of (a, b) are also greatest common divisors and any gcd's are associates. Assume a unit u and its inverse v, along with some other unit k (and inverse k*): there is a gcd of (a, b) d with d* = dk = au* + bv*, then d = d*k* = (au* + bv*)k* = a(u*k*) + b(v*k*) = au + bv. This provides a form for the gcd.

Take a, b, c ∈ R with a | bc and a, b to be relatively prime. Then ∃n : an = bc, but since (a, b) = 1, then a | c. The definition of irreducibles does not change, however, the factorization of nonzero, nonunit elements in R are now only unique up to associates. The proof of this follows standard procedure, but as it's a little unintuitive, I'll go through it once here. We want to show that the set S of all nonzero nonunit elements of R that are not products of irreducibles is empty. This defines a subset with at least one factorization (as otherwise it would be a unit) {f(s) | s ∈ S}, where f(s) are non-negative with a smallest element by the Well-Ordering Axiom, which holds, since Integral domains retains ordering from the ring axioms. This smallest element f(a) can't be irreducible, so a = bc however, neither b nor c is a unit. Of course f(b) ≤ f(bc) = f(a), so f(a) couldn't be the smallest element in S or f(b) is not in S. Either way, this leads to a contradiction and concludes the proof.

Principal Ideal Domains (PID) are integral domains in which every ideal is principal. We construct an ideal in an Euclidean domain, to show that all Euclidean domains are principal ideal domains. Some nonzero ideal I in the Euclidean domain R induces the set of non-negative integers {f(i) : i ∈ I} with some smallest element by the Well-Ordering Axiom. I is the principal ideal (b) = {rb : r ∈ R}, and as b ∈ I, rb ∈ I for all r ∈ R. For the other direction, suppose c ∈ I, then there are some q, r ∈ R: c = bq + r and r = 0. Otherwise f(r) < f(b). Since r = c - bq and c, b ∈ I, then r ∈ I. This leads to a contradiction and all ideals in R must be principal.

At this point, it's probably wise to include some words on principal ideals in integral domains anyway. Let's take two principal ideals (a) ⊂ (b). This means that a ∈ (b), but since (b) only includes multiples of b, then b | a. If we additionally want that a | b, then (a) = (b). Conversely, if we want (a) ⊂ (b), but expressly not (a) = (b), then expressly not a | b. Since b and a are associates iff a | b and b | a, then a and b are never associates. We can then of course construct a chain of inclusions of principal ideals. We call that the ascending chain condition (ACC), a condition that all PID's satisfy, due to each inclusion automatically leading to this condition by multiplication of elements. Using this, one can get a the same argument as the one we concluded the section about Euclidean domains with, arguing that every nonzero, nonunit element of a PID is a product of irreducibles with a factorization that is up to associates. From this property, we construct a unique factorization domain (UFD) consisting only of these elements. Clearly, PID's are a subset of UFD's.

For any a, b in a UFD R there are units u, v and irreducibles p(i) of which none are associates of one another such that

Of course both c and d have factorizations into products of irreducibles, some of which may have associates in R. These can be written as a unit, leaving only irreducibles without associates and one unit. Any UFD satisfies the ACC through this presentation. For irreducible elements p in UFD's we can say that if p | bc, then (assuming a non-trivial situation), then if c is a unit, then pt = bc, and it follows p | b. Analogue for the case that b is a unit. If neither are nonunits, then we can use the representation of products of irreducibles. pt = bc, which is equal to the product of the product presentations of b and c, so p must be the associate of one of these irreducibles. It then divides either b or c, depending on which of the two, the associate is a factor of. This is also true in reverse. If the statement is true for all p in R and R satisfies the ACC, then R is a UFD.

Assume d is a gcd of a set of elements in R, then d's associates are also gcd's following the same logic utilized at the beginning of the chapter. In fact, any two gcd's of a set in R are associates, since a gcd divides all elements of a set, which gives the satisfies the definition of gcd's of the other. Since every nonzero element in a UFD has a product representation of non-associate irreducibles and a unit, one can collect the smallest exponent of each irreducible and take the product of those. This returns a gcd of all elements.

We're already familiar with factorization in integral domains through factorization in Z. Similarly, we know that for some constructions of polynomial solutions, we need to use elements that are not exactly part of the domain the polynomial is supposed to live in. This is where the square root is usually defined. We define square-free integers as integers with no squares (besides 1) in their factorization and the norm as N(s+t√(d)) = (s + t√d) (s - t√d). We can see that N(a) = 0 iff a = 0 and N(a)N(b) = N(ab) trivially. Let d be square free and u ∈ Z[√d] a unit, then N(u)N(u*) = N(1) = 1 -> N(u) = ± 1. Conversely, if u = s+t√(d) and N(u) ± 1, then uu* = N(u) = ± 1, and u(± u*) = 1 means that u is a unit. If d = -1, then we can check that the units in Z[√(d)] are {± 1, ± i}, and if d < -1, Z[(d)]] they are {± 1}. However, if d > 1, then Z[√d] has infinitely many units. Now we can check what happens when the norm evaluates to a prime. Assume p ∈ Z[√d] and N(p) ≠ ± 1, then p is not a unit in Z[√d]. If p = ab, then N(p) = N(a)N(b) -> N(a) = ± 1 or N(b) ± 1. One of the two factors then must be a unit and p is irreducible. Every nonzero, nonunit element N(a) in Z[√d] is a product of irreducible elements by the same chain of logic used in the past for this kind of proof.

At this point, it might be wise to define the product of ideals. Ideals are a ordered set of elements, so the product of ideals is conveniently defined as the set, where the elements of both ideals with the same index are multiplied, and their sum taken. At this point we also introduce the algebraic number, a complex number that is the root of monic polynomial with rational coefficients. For the algebraic number t in the domain of all algebraic integers in Q(t) R, every non-trivial ideal of R is the product of prime ideals unique up to the order of factors.

We can extend the definition of some UFD R to R[x], the space of all polynomials in R. In R[x], every nonzero, nonunit f(x) is a product of irreducible polynomials by the same logic that applies to all other polynomial spaces. It also retains its elements' properties regarding primes. If the only constants dividing a polynomial in R[x] are the units in R, then it's considered primitive. The product of such polynomials also come out primitive, because the set of constant divisors are products of the constant divisors that divide both factor polynomials. If there are some nonzero elements r, s in R, and f(x), g(x) primitive polynomials in R[x], such that rf(x) = sg(x), then f(x) = r*sg(x) and since r*s divides f(x), then it's a unit (with an element we'll call u). Since u is also a unit, then su = r, so r and s are associates in R. Similarly, through representation of r, s as products of irreducibles, f(x) and g(x) are associates in R[x].

If f(x) and g(x) are primitive and associate in R[x]'s field of quotients F[x], then g(x) = r/s f(x) = kf(x) for some nonzero k. Then sg(x) = rf(x) in R[x], and f(x) is associate to g(x). Let's assume f(x) has a positive degree and is irreducible in R[x]. If f(x) is not irreducible in F[x], then it has a product presentation f(x) = g(x)h(x) with positive degree. If b has a lcd of the coefficients in g(x), then bg(x) = aj(x) with j(x) a primitive polynomial with positive degree. The same with h(x). Then f(x) = (ac)(bd)j(x)k(x) in R[x], which is a contradiction, so f(x) must also be irreducible in F[x]. Through inspection of coefficients, we can conclude that if R is a UFD, then R[x] is too.

All of the preceding paragraph can be used to show that Z[x] is a UFD, but not a PID.

Field Extensions

Before getting to the whole point of this exercise: The Galois Theory, we have to first take a look at a field can be extended in an algebraically correct manner. We call K a finite-dimensional extension of F, if K is finite dimensional over F. We write the extension order as [K:F]. From that follows intuitively that if [K:F] = 1 iff K = F. Assume that F ⊂ K ⊂ L and [K:F] = m, [L:K] = n finite. Then there are bases of size m for K over F and of size n for L over K. The elements of such bases are trivially nonzero as to retain the required degrees of freedom in the space, so their products are also nonzero. As the basis is minimal, the vectors are linearly independent from one another. There are exactly mn unique products of vector products with the first out of U = B([K:F]) and the second out of V = B([L:K]). Let's call the set of these products W.

Any element l ∈ L is a linear combination of elements in V with coefficients in K, and every k ∈ K is a linear combination of elements in U with coefficients in F. After substitution, every element in L can be written as a linear combination of elements in V, K and F. The set of all products of basis vectors span L over F. Linear independence of this basis follows from linear independence of the constructing sets, which were also bases. Then, [L:F] = [L:K][K:F].

Finite dimensional extension fields K, L of F and an isomorphism f:K ⟶ L with f(c) = c ∀ c ∈ F. [K:F] = n and nonzero with some basis U with n elements. f is an isomorphism, so we can map f(u ∈ U) ⟶ v ∈ V = B([L:F]). It definitely is a basis due to f(ax + by) = f(a)f(x) + f(b)f(y), for a, b ∈ F. However, f(a) = a and f(b) = b by assumption, so f(v) spans L.

We can define another field through an intersection of fields. For example, an extension field K of F with k ∈ K. We write F(k) for the subfields in K containing F and k (it contains at least K). As an intersection of fields, F(k) is a field and contained with every subfield of K containing F and k. It's by definition the smallest subfield containing both. We call F(k) a simple extension of F. k is algebraic over F, if u is the root of some nonzero polynomial in F[x]. An element of K that is not a root of some non-trivial polynomial in F[x] is transcendental in F.

Assume an algebraic element u that is a the root of two monic irreducible polynomials p(x) and q(x). Irreducibility follows the same proof as always. Assume u is also the root of g(x). g(x) is not monic, and not irreducible, so g(x) = p(x)s(x) + r(x) by division algorithm. Since u is a root of both, r(u) = g(u) - p(u)s(u) = 0 + 0 = 0. Now we take a look at q(x). It shares u as a root of p(x), then p(x) | q(x), and q is irreducible, meaning that q(x) = cp(x), c ∈ F. Since q(x) was supposed to be monic, then c = 1. This means that the monic irreducible polynomial with root u is unique for each u.

If p(x) is minimal with degree n instead, then F(u) contains F, and with it all its elements that takes the polynomial form f(u). g ∈ F[x] is in the kernel of ϕ : F[x] ⟶ F(u), ϕ(f(x)) = f(u), iff ϕ has u as a root. The kernel of ϕ is the principal ideal (p(x)). The First Isomorphic Theorem gives an isomorphism between F[x]/(p(x)) and Im ϕ. Due to irreducibility, F[x]/(p(x)) and Im ϕ are both fields and every constant gets mapped to itself. There is some x so that ϕ(x) = u, so Im ϕ is a subfield of F(u) containing F and u. Since F(u) is also supposed to be the smallest subfield containing both those elements, F(u) = Im ϕ. It also follows that every v ∈ F(u), v ≠ 0 can be written as f(v) for some function f in F[x]. If the degree of p(x) = n, then f(x) can be written in terms of p(x) through division algorithm, so that f(u) is a linear combination of coefficients and powers of u, meaning that those n first powers of u (including 0) spans F(u). They then are a basis of it and [F(u) : F] = n.

Algebraic extensions are extensions containing only algebraic elements over the extended set. Finite-dimensional extension fields K over F are algebraic. Finite-dimensional extension fields have finite bases |B| over F, which spans K. That means that every linearly independent set in K has at most |B| elements. Either:

Or all non-negative powers of u are distinct, in which case a basis of n + 1 elements can be constructed of the subsequent powers of u starting from 0. These are trivially linearly independent over F and

u is trivially a root of the sum, which also trivially lives in F[x], but as u was defined in K, not in F, then the sum is transcendental in F. A field K can be obtained by a chain of extensions F({u}), each adding an algebraic element into the extension. That is, because the extension can be treated similarly to an isomorphy, i.e. F(a, b) = F(a)(b). Let a, b be linearly independent algebraic integers over F, then they are each roots to different polynomials in F. Then, b is also algebraic in F, and vice versa. Simple extensions by algebraic elements are finite-dimensional by previous proofs and so

is finite for all t > 1. [K:F] can then be written as successive products of field extensions.

This makes [K:F] finite and K algebraic over F. This also implies that algebraic extension fields can be chained as well, so if L is an algebraic extension field of K and K an algebraic extension field of F, then L is automatically an algebraic extension of F. The set E of K containing all elements that are algebraic over F, then for all e, f ∈ E are algebraic over F, and F(e, f) is an algebraic extension by definition. By closure of operations in F(e, f), it's a subgroup of E and E is a field. E then fulfills the requirements to be a subfield of K.

Extension fields K over F, in which nonconstant polynomials f(x) can be completely factored in K[x] as a product of first order polynomials up to some scaling factor, are said to contain all roots and we define f(x) as splitting over K. K is a splitting field, if the factored f(x) contains exactly one first-order polynomial factor for each algebraic extension element. If f(x) has a degree n in F(x), then there exists a splitting field L of f(x) over F, with [L:F] ≤ n!. This follows from construction of splitting fields, and through induction over the degree of f(x). Splitting fields of the same polynomial are also isomorphic, though there is a stronger statement that is easier to prove.

Let α : F → E an isomorphism and f(x) nonconstant in F[x]. αf(x) then is the corresponding polynomial in E[x]. A splitting field K of f(x) over F and a splitting field L of αf(x) over E has α act as an isomorphism between K and L. To show this,we can map the chains of E ⊆ E(v) ⊆ L and F ⊆ F(u) ⊆ K onto one another, as we know that α maps F onto E. F(u) will map isomorphically onto E(v) via roots u of some polynomial p(x) in K and v of α p(x) in L. By induction hypothesis, as α acts isomorphically, K is isomorphic to L.

Some splitting field K of f(x) ∈ F[x], then we have seen that K is the full extension of F with all roots of f(x). An irreducible polynomial p(x) in F[x] with a root v ∈ K and a splitting field L over K, then we can chain these fields together like this: F ⊆ K ⊆ L. Any root w of p(x) in L constructs isomorphisms between F(v) and F(w). K(w) is then a splitting field of f(x) over F(w) and so K also splits over F(v). This defines element-wise an isomorphism K → K(w), mapping v to w, and all elements in F to themselves. [K:F] = [K(w):F] and [K:F] = [K(w):F] = [K(w):K][K:F] with [K(w):K] = 1 → K(w) = K, meaning that w ∈ K. K is then normal over F. This is also true the other way around, shown formally through splitting over products of algebraic equations.

The phenomenon of repeated roots (or roots of higher order) introduces separability to the classification of polynomials. A polynomial is separable, if it has no repeated roots in any splitting field. We assume that the derivation of polynomials is a known concept. Separability can also be determined from within the field F[x] in which a polynomial f(x) lives in. If it is prime with it's derivative, then f(x) is prime. Assume f(x) is not separable, and let K be it's splitting field. f(x) has some repeated root u in K, which factors f(x) = (x-u)(x-u)g(x). The derivative then is (x-u)(x-u)g'(x) + 2(x-u)g(x) and f'(u) = 0. As u is also a root f(x), they are not relatively prime.

Fields have characteristic 0, if n ≠ 0 ∀ positive n. In polynomial fields of such fields, irreducible polynomials p(x) ∈ F[x] have nonzero derivatives, which they are relatively prime with. They are then all separable and every algebraic extension is separable. Such extensions become simple, if they are finite-dimensional.

Integral domains R have characteristics of either 0 or some positive prime. We understand characteristic 0, so we assume that the characteristic is n ≠ 0. If n were not prime, then we ∃ k, t: n = kt, k < n and t < n. Then, (k1)(t1) = n1 = 0. This is an issue for integral domains. Rings with identity inherit this trait. Rings R with identity and characteristic n > 0 and k1 = 0, then n | k. This is trivial. So is the converse, after application of the division algorithm. From this, we can gather all the properties and conclude that every finite field has some prime as characteristic, as it's not infinite by definition and inherits the integral domain characteristic traits. Finite fields also have orders of the n-th power of their characteristic p and n = [K : Z/p]. This is a very handy trait that will make classifying finite fields a lot easier. We can even use it to equate the same order for an extension field of Z/p, iff K is a splitting field of x^q - x over Z/p, where q is their order. Many proofs applied to the extension fields carry over, such as the isomorphism of finite fields of the same orders and the simple extension of subfields. Another important consequence is the existence of an irreducible polynomial of any positive degree in Z/p[x] for all primes p.