## Galois Theory: Field Extensions

I’m going to write the next few math posts on Galois theory, which I handwaved into my algebraic number theory posts. At the center of Galois theory is the idea of field extensions, which I’ll consistently denote as L/K.

L naturally inherits a vector space structure over K. So it makes sense to speak of a dimension, which is written [L:K]. The case of interest in elementary Galois theory is when [L:K] is finite, in which case the extension is called finite. It’s possible to talk about infinite Galois theory too, though, so I’ll make as much as possible in this series independent of the finiteness of dimension.

If M/L/K is a tower of field extensions, then [M:K] = [M:L][L:K]. To see why, let n1 = [M:L] and n2 = [L:K]. If m1, m2, …, m(n1) form a basis for M/L, and l1, l2, …, l(n2) form a basis for L/K, then every m in M can be written as l‘1m1 + … + l‘(n1)m(n1) = k11l1m1 + k12l2m1 + … + k(1, n2)l(n2)m1 + … + k(n1, n2)l(n2)m(n1), where the k(i, j)’s are in K, making the l(j)m(i)’s a spanning set of size n1n2. It’s trivial to show that any linear dependence on this spanning set induces a linear dependence on the m(i)’s or the l(j)’s, which contradicts the fact that they’re bases.

A field F is called an intermediate field of L/K if it’s a subfield of L that contains K. As a corollary to the above tower law, [F:K][L:F] = [L:K] so that [F:K] and [L:F] divide [L:K] if it’s finite.

If a is an element of L, then a is called algebraic over K if it is the root of some polynomial over K. Otherwise, it’s called transcendental. If L = C and K = Q, then i, SQRT(2), and 3 + 2^(1/17) are algebraic, while pi and e are transcendental. If every element of L is algebraic over K, we call L/K an algebraic extension.

If L/K is finite, then it must be algebraic. To see why, let [L:K] = n, and let a be any element of L. Then the set {1, a, a^2, …, a^n} has n + 1 elements, so it’s linearly dependent, and there exist numbers k0, k1, …, k(n) in K not all zero such that k(n)a^n + … + k1a + k0 = 0. That generates a polynomial over K that has a as a root, so a is algebraic over K.

If L/K is a field extension and a is an element of L, then we write K(a) for the smallest intermediate field containing a, and K[a] for the smallest subring of L containing K and a.

We can form the ring K[x] of polynomials in one variable, x, with coefficients in K. That ring has fraction field K(x), the field of all rational functions in one variable over K. Now, K[x] projects onto K[a] by sending x to a. The resulting ring homomorphism has nonzero kernel iff there exists a nonzero polynomial over K, f, such that f(a) = 0, which holds iff a is algebraic over K.

So if a is transcendental, then the homomorphism has a nonzero kernel, and K[a] is isomorphic to K[x] so that K(a) is isomorphic to K(x). If a is algebraic, then K[a] is isomorphic to some quotient of K[x]. The kernel of the map from K[x] to K[a] is then a nonzero ideal, so as I showed ages ago, it’s generated by a single polynomial, the lowest degree polynomial that has a as a root.

Now, that polynomial is unique up to multiplication by a unit, i.e. scaling by some element of K, so it’s convenient to scale it to have leading coefficient 1. In that case it’s called the minimal polynomial of a over K and denoted by m(a)(x), or m(a), where the a is supposed to be a subscript but is written here in parentheses for ASCIIfication purposes.

Further, m(a) is an irreducible polynomial. Otherwise, write m(a) = fg with deg(f), deg(g) > 0, and note that since m(a)(a) = 0, either f(a) or g(a) is 0. But that contradicts the choice of m(a) as the polynomial of least degree that has a as a root. Therefore, K[a] = K[x]/(m(a)(x)) is a ring modulo a maximal ideal, so it’s a field, and K(a) = K[a].

Also, [K(a):K] = deg(m(a)) when a is algebraic. The proof is fairly straightforward: if deg(m(a)) = n, then {1, a, a^2, …, a^(n-1)} is linearly independent by the minimality of the degree of m(a). It spans K(a) = K[a] because it spans a^n by using m(a), a^(n+1) by using x*m(a)(x) and {a, a^2, …, a^n}, etc.; then {1, a, a^2, …} spans K[a].

To recap, a is algebraic over K iff the following equivalent conditions hold:

1. a is a root of a polynomial with coefficients in K.
2. [K(a):K] is finite.
3. K(a) = K[a].
4. K[a] is a proper quotient of K[x].

The important properties are 1 and 2.

If E and F are two intermediate fields of L/K, then the smallest intermediate field containing both E and F is denoted EF. Then [EF:E] <= [F:K] and [EF:F] <= [E:K], so [EF:K] <= [E:K][F:K]. To show that, it’s enough to prove that [EF:E] <= [F:K].

But if b1, b2, …, b(n) are basis elements for F/K, then they span EF/E. Closure under addition and subtraction is obvious, and closure under multiplication and division follows from writing each b(i)b(j), which is in F, as a sum k1b1 + … + k(n)b(n). If they also form a basis, then [EF:E] = [F:K], [EF:F] = [E:K], and [EF:K] = [E:K][F:K], in which case E and F are called linearly disjoint over K. If they don’t, then the inequalities are all strict.

A necessary but not sufficient condition for E and F to be linearly disjoint over K is that their intersection is K; to see that it’s not sufficient, let K = Q, L = Q(2^(1/4), i), E = Q(2^(1/4)), F = Q(i2^(1/4)), where EF = L, [L:K] = 8, [E:K] = [F:K] = 4. I’ll come back to a sufficient condition later on, after developing Galois theory more.

For now, note that if a and b are two algebraic elements of L/K, then [K(a)K(b):K] <= [K(a):K][K(b):K] which is finite, so every algebraic expression in a and b is algebraic over K. In particular, a + b, ab, ab, and a/b are algebraic over K, so that the set of all algebraic elements of L/K forms an intermediate field, called the algebraic closure of K in L.

Even infinite Galois theory deals only with algebraic extensions, albeit possibly infinite ones, which exist (e.g. this convoluted example of Q(SQRT(2), SQRT(3), SQRT(5), …)).

Finally, every field has an invariant called its characteristic. If 1 + 1 + … + 1 is always nonzero in a field K, we say that the characteristic of K is 0; then K contains N, so it contains Z, so that it contains its fraction field, Q.

If 1 + 1 + … + 1 = 0 after n steps but not after any smaller number of steps, we say that char(K) = n. In fact n is prime, because if n = ab, then (1 + 1 + … (a times) + 1)(1 + 1 + … + (b times) + 1) = 0 and neither factor is 0, which implies that both factors are nonzero but have no multiplicative inverse.

For each prime p, if char(K) = p then K contains F(p) = Z/pZ by adding successively many copies of 1 to itself. In each case, that field K must contain is called its prime subfield. Classical Galois theory always has K = Q; it’s also possible to build Galois theory from F(p), but finite fields’ finite extensions are remarkably easy to understand.