A splitting field of a polynomial p(X) over a field K is a field extension L of K over which p factors into linear factors
Given an algebraically closed field A containing K, there is a unique splitting field L of p between K and A, generated by the roots of p. If K is a subfield of the complex numbers, the existence is immediate. On the other hand, the existence of algebraic closures in general is often proved by 'passing to the limit' from the splitting field result, which therefore requires an independent proof to avoid circular reasoning.
Given a separable extension K′ of K, a Galois closure L of K′ is a type of splitting field, and also a Galois extension of K containing K′ that is minimal, in an obvious sense. Such a Galois closure should contain a splitting field for all the polynomials p over K that are minimal polynomials over K of elements a of K′.
Finding roots of polynomials has been an important problem since the time of the ancient Greeks. Some polynomials, however, such as x2 + 1 over R, the real numbers, have no roots. By constructing the splitting field for such a polynomial one can find the roots of the polynomial in the new field.
The irreducible factor fi(X) used in the quotient construction may be chosen arbitrarily. Although different choices of factors may lead to different subfield sequences, the resulting splitting fields will be isomorphic.
As mentioned above, the quotient ring Ki+1 = Ki[X]/(f(X)) is a field when f(X) is irreducible. Its elements are of the form
where the cj are in Ki and α = π(X). (If one considers Ki+1 as a vector space over Ki then the powers αj for 0 ≤ j ≤ n−1 form a basis.)
The elements of Ki+1 can be considered as polynomials in α of degree less than n. Addition in Ki+1 is given by the rules for polynomial addition and multiplication is given by polynomial multiplication modulo f(X). That is, for g(α) and h(α) in Ki+1 their product is g(α)h(α) = r(α) where r(X) is the remainder of g(X)h(X) divided by f(X) in Ki[X].
The remainder r(X) can be computed through long division of polynomials, however there is also a straightforward reduction rule that can be used to compute r(α) = g(α)h(α) directly. First let
The polynomial is over a field so one can take f(X) to be monic without loss of generality. Now α is a root of f(X), so
If the product g(α)h(α) has a term αm with m ≥ n it can be reduced as follows:
Consider the polynomial ring R[x], and the irreducible polynomial x2 + 1. The quotient ring R[x] / (x2 + 1) is given by the congruence x2 ≡ −1. As a result, the elements (or equivalence classes) of R[x] / (x2 + 1) are of the form a + bx where a and b belong to R. To see this, note that since x2 ≡ −1 it follows that x3 ≡ −x, x4 ≡ 1, x5 ≡ x, etc.; and so, for example p + qx + rx2 + sx3 ≡ p + qx + r⋅(−1) + s⋅(−x) = (p − r) + (q − s)⋅x.
The addition and multiplication operations are given by firstly using ordinary polynomial addition and multiplication, but then reducing modulo x2 + 1, i.e. using the fact that x2 ≡ −1, x3 ≡ −x, x4 ≡ 1, x5 ≡ x, etc. Thus:
If we identify a + bx with (a,b) then we see that addition and multiplication are given by
We claim that, as a field, the quotient R[x] / (x2 + 1) is isomorphic to the complex numbers, C. A general complex number is of the form a + bi, where a and b are real numbers and i2 = −1. Addition and multiplication are given by
If we identify a + bi with (a,b) then we see that addition and multiplication are given by
The previous calculations show that addition and multiplication behave the same way in R[x] / (x2 + 1) and C. In fact, we see that the map between R[x]/(x2 + 1) and C given by a + bx → a + bi is a homomorphism with respect to addition and multiplication. It is also obvious that the map a + bx → a + bi is both injective and surjective; meaning that a + bx → a + bi is a bijective homomorphism, i.e. an isomorphism. It follows that, as claimed: R[x] / (x2 + 1) ≅ C.