Galois Theory: A Study of Cyclotomic Field ... - Scripps College

8 Galois Theory
symmetry [A, p.538]. Let us go back to our example of the polynomial
x 2 + 1 = 0. We already know that our polynomial has no roots in Q, but it
does have roots in C. Complex conjugation, the first model for the symmetric
relationships we will be studying, gives us the roots ±i and leaves the
rationals fixed. Any quadratic (degree 2) field extension will have similar
symmetries.
As an example, let us look at a degree 2 extension K/F generated by
any α ∈ K which is not in F . Let f(x) = x 2 + bx + c be a polynomial with
coefficients in F such that α is a root of f(x). We see that, b = −(α + α ′ ), so
α ′ = −(α + b) is the other root. Over K, the polynomial f(x) splits into the
linear factors (x − α)(x − α ′ ).
Our symmetry comes from the fact that both α and α ′ are roots of f(x).
Proposition 2. Let L = F (α) and L ′ = F (α ′ ) be two extension fields of F
generated by the algebraic elements α and α ′ . Then there is an isomorphism of
fields
σ : F (α) ˜→F (α ′ ),
which is the identity on F and sends α → α ′ if and only if the irreducible polynomials
for α and α ′ over F are equal.
Proof. See [A, p.496]
According to the above proposition,
σ : F (α) −→ F (α ′ )
is an isomorphism which sends α → α ′ and is the identity on the field F .
In our case, however, K can be generated by either of the two roots. That