Example: 
Say f(x) in Q[x], with rational coefficients. If the degree is n, then it has n complex roots (with multiplicity). 

The smallest subfield of C that contains all those roots is called the “splitting field” of f.
(Because f(x) = a_0 (x - alpha_1) … (x - alpha_n) splits into linear factors.)

Q(alpha_1, …, alpha_n) is created in steps
Q(alpha_1) 
Q(alpha_1, alpha_2) = Q(alpha_1)(alpha_2)
Etc.

How can we do this for arbitrary fields?

Lemma 1. (Creating roots algebraically)
Let k be a field, let f(x) be an irreducible polynomial in k[x]. Then

E = k[x]/(f(x))

is a field extension of k, and f(x) has in root in E.

Proof. 
f(x) is irreducible, so ideal (f(x)) is a maximal ideal. Therefore, the quotient ring is a field.
E contains a copy of k:

k[x] —> E, g(x) —> [g(x)] = g(x) + (f(x))

k —> E, a —> [a] is injective, so it embeds k into E. So k contained in E is a field extension.

Set alpha = [x] in E. This is a root. Reason: If we write

f(x) = a_n x^n + … + a_1 x + a_0,

then f([x]) = a_n [x]^n + … + a_1 [x] + a_0 = [f(x)] = [0] in E.   Q.E.D.

We can iterate this procedure to factor any polynomial into linear factors.

Proposition 2.
Let k be a field, f(x) in k[x] a polynomial. Then there is a field extension E of k in which f(x) factors completely into linear factors. (So if deg f(x) = n, it has exactly n roots in E.)

Proof. 
By induction on n = deg f(x).
k[x] has unique factorization, so we can write f(x) as a product of irreducible polynomials.
Pick one of the irreducible factors, say p_1(x), and we use Lemma 1 to create a first field extension E_1 in which p(x) has a root: E_1 = k[x]/(p_1(x)). 
Now we consider f(x) as a polynomial with coefficients in E_1, and we factor it into irreducible polynomials. If they are all of degree 1, we are done; if not, then we take one irreducible factor, say p_2(x), and create a new field extension E_2 of E_1 in which p_2(x) has a root: E_2 = E_1[x]/(p_2(x)). And so on.
At each step, the product of those irreducible factors that are not linear gets smaller (because we create one extra root). So by induction, we are done. Q.E.D.

Definition: 
Let f(x) be a polynomial in k[x]. A field extension E of k is called a “splitting field” for f(x) if 
(1) f(x) splits into linear factors in E[x], i.e. if deg f(x) = n, then f(x) has n roots in E.
(2) E is minimal with this property, i.e. f(x) does not factor into linear factors over any subfield of E. 

Example: 
There are many different splitting fields: for instance, if f(x) in Q[x], then we have constructed a splitting field inside C; but we can also use the abstract construction from above to create another splitting field. 

Any polynomial has a splitting field, by the construction above: construct a field extension in which f(x) has enough roots, and then take the smallest subfield containing all those roots. 

Example:
f(x) = x^2 - 3 in Q[x]
One splitting field is Q(sqrt(3)) inside C. 
Another splitting field is E = Q[x]/(x^2-3). 

They are isomorphic as fields: Q[x] —> C, x —> sqrt(3).
Minimal polynomial is x^2-3, so kernel is (x^2-3), so

E = Q[x]/(x^2-3) —> Q(sqrt(3))

is an isomorphism. 

Theorem 3. 
Let f(x) be in k[x]. Then any two splitting fields of f(x) are isomorphic as fields.
More precisely, if E and E’ are two splitting fields, then there is a field isomorphism

phi : E —> E’ 

whose restriction to the subfield k is the identity (i.e. for any a in k, phi(a) = a).

(So we can talk about “the” splitting field, at least up to isomorphism.) 

Idea: We start from the identity isomorphism from k to k, and then we extend it step by step to an isomorphism from E to E’ 

Observation: If E is a splitting field of f(x), then E is a finite extension of k. 
(Reason: f(x) has finitely many roots, and the subfield of E generated by those roots is a finite extension. By (2) in definition, E must be that subfield.)
Point: dim_k(E) is finite.

Proof.
We have the isomorphism id: k —> k. We are looking for an extension of this to an isomorphism phi: E —> E’. 

Consider pairs (F, psi), where F is a subfield of E containing k: k inside F inside E
and where psi : F —> E’ is a ring homomorphism that extends id: k —> k. 

We partially order these in the obvious way: (F, psi) <= (F’, psi’) if F’ contains F and the restriction of psi’ to F is equal to psi. i.e. if psi’ extends psi from the smaller field F to the bigger field F’. 

Pick a maximal element: concretely, pick a pair (F, psi) in which dim_k(F) is as big as possible. 

Claim: F = E.
Suppose that F is not equal to E. Since E is generated by the roots of f(x), this means that at least one of the roots of f(x) is not contained in F. Factor f(x) into irreducible polynomials in F[x], and let p(x) be one of the factors that is not linear. Let alpha be one of the roots of p(x) inside E. 
Map everything to E’ by using psi: Set F’ = psi(F), so that psi : F —> F’ is an isomorphism.
Let p’(x) = image of p(x) under the field isomorphism psi: F —> F’. 
If p(x) = a_r x^r + … + a_1 x + a_0, then 
p’(x) = psi(a_r)  x^r + … + psi(a_1) x + psi(a_0) in F’[x]. 
Since p[x] in F[x] is irreducible, and since psi : F —> F’ is an isomorphism of fields,
the new polynomial p’(x) in F’[x] is still irreducible. Also, p(x) divides f(x), and so p’(x) divides f(x) (because f(x) has coefficients in k, and psi is the identity on k.) 
But f(x) also factors into linear factors in E’, and so p’(x) has a root in E’. Call this alpha’. 

We can extend F to a bigger field: F(alpha) inside E, isomorphic to F[x]/(p(x)). 
We can also extend F’ to a bigger field: F’(alpha’) inside E’, isomorphic to F’[x]/(p’(x)). 

We can extend psi : F —> F’ to an isomorphism between those bigger fields:
mu : F[x]/(p(x)) —> F’[x]/(p’(x))
mu(x) = x, on the coefficients, mu is equal to psi. 

So we get a pair (F(alpha), mu) that is strictly bigger than (F,psi). 

This is a contradiction; the only way to avoid it is that actually F = E. 

For dimension reasons, we get that psi : E —> E’ an isomorphism. Q.E.D. 

From now on, we talk about “the” splitting field, keeping in mind that it is unique up to isomorphism. 

Extended example: finite fields

Reminder: characteristic of a field k

 char(k) = p if 1 + …. + 1 = 0 (sum has p terms)
 char(k) = 0 if this doesn’t happen. 

Interpretation: consider the ring homomorphism f: Z —> k
f(0) = 0,
f(n) = 1 + …. + 1 (n times) for n > 0
f(-n) = (-1) + … + (-1) (n times) for n > 0.

The kernel is a prime ideal, so either (0) or (p) for some prime number p.
If the kernel is (p), we get Z/(p) —> k, so we get a copy of the field F_p = Z/(p) with p elements inside k.
If the kernel is (0), Z embeds into k, therefore Q embeds into k
Q —> k, a/b —> f(a)/f(b) (okay since f(b) nonzero for b nonzero).

char(k) = p   equivalent to k contains a copy of F_p
char(k) = 0  equivalent to k contains a copy of Q.

Let E be a finite field, meaning with finitely many elements. Therefore char(E) = p prime.
So we have F_p inside E, and E must have finite dimension as a F_p-vector space. 
Say n = dim of E over F_p, then E has p^n elements. 

We want to construct all possible finite fields. 

Observation. Say #E = p^n. Then E* = E - {0} is a group under multiplication, which has p^n-1 elements. By Lagrange’s theorem (about groups), any element a in E* satisfies

a^(p^n-1) = 1.

So every element of E is a root of the polynomial

f(x) = x^(p^n) - x in F_p[x].

We can use this observation to construct finite fields. Fix a prime p, and an integer n >=1.
Let E be the splitting field of polynomial x^(p^n) - x, as a field extension of F_p. Unique up to isomorphism (by Theorem 3).

Claim: #E = p^n. 

Reason: All the roots of f(x) = x^(p^n) - x are distinct, so they are exactly p^n of them. 
In fact, E is exactly the set of these p^n roots. 

Derivative f’(x) = p^n x^(p^n-1) - 1 = -1 (p = 0 in F_p), this has no common factor with f(x). Therefore f(x) has p^n distinct roots in E. (Details: next time)

In characteristic p, the set of roots actually forms a field: if a^(p^n) = a, and b^(p^n) = b,
Then the same thing is true for a+b and ab

(ab)^(p^n) = a^(p^n) b^(p^n) = ab
(a+b)^(p^n) = a^(p^n) + b^(p^n) = a +b. 
(All the binomial coefficients p^n choose k are divisible by p, hence 0 in characteristic p.)

Conclusion: For every prime p and every n>=1, there is (up to isomorphism) a unique finite field with p^n elements. Concretely, it’s the splitting field of x^(p^n) - x.