# Splitting field

Splitting field

In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial factors (or "splits", hence the name) into linear factors.

## Definition

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

$p(X) = \prod_{i=1}^{\deg(p)} (X - a_i) \in L[X]$

and such that the coefficients ai generate L over K. The extension L is then an extension of minimal degree over K in which p splits. It can be shown that such splitting fields exist and are unique up to isomorphism. The amount of freedom in that isomorphism is known to be the Galois group of p (if we assume it is separable).

## Facts

An extension L which is a splitting field for multiple polynomials p(X) over K is called a normal extension.

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 automatic. On the other hand, the existence of algebraic closures in general is usually proved by 'passing to the limit' from the splitting field result; which is therefore proved directly 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′.

## Constructing splitting fields

### Motivation

Finding roots of polynomials has been an important problem since the time of the ancient Greeks. Some polynomials, however, have no roots such as X2 + 1 over $\mathbb R$, the real numbers. By constructing the splitting field for such a polynomial one can find the roots of the polynomial in the new field.

### The Construction

Let F be a field and p(X) be a polynomial in the polynomial ring F[X] of degree n. The general process for constructing K, the splitting field of p(X) over F, is to construct a sequence of fields $F=K_0, K_1, \ldots K_{r-1}, K_r=K$ such that Ki is an extension of Ki − 1 containing a new root of p(X). Since p(X) has at most n roots the construction will require at most n extensions. The steps for constructing Ki are given as follows:

• Factorize p(X) over Ki into irreducible factors $f_1(X)f_2(X) \cdots f_k(X)$.
• Choose any nonlinear irreducible factor f(X) = fi(X).
• Construct the field extension Ki + 1 of Ki as the quotient ring Ki + 1 = Ki[X] / (f(X)) where (f(X)) denotes the ideal in Ki[X] generated by f(X)
• Repeat the process for Ki + 1 until p(X) completely factors.

The irreducible factor fi 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.

Since f(X) is irreducible, (f(X)) is a maximal ideal and hence Ki[X] / (f(X)) is, in fact, a field. Moreover, if we let $\pi : K_i[X] \to K_i[X]/(f_1(X))$ be the natural projection of the ring onto its quotient then

$f(\pi(X)) = \pi(f(X)) = f(X)\ \bmod\ f(X) = 0$

so π(X) is a root of f(X) and of p(X).

The degree of a single extension [Ki + 1:Ki] is equal to the degree of the irreducible factor f(X). The degree of the extension [K : F] is given by $[K_r : K_{r-1}] \cdots [K_2 : K_1][K_1 : F]$ and is at most n!.

### The Field Ki[x]/(ƒ(x))

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 $c_{n-1}\alpha^{n-1} + c_{n-2}\alpha^{n-2} + \cdots + c_1\alpha + c_0$ 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 1 ≤ jn−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 the product 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 $f(X) = X^n + b_{n-1} X^{n-1} + \cdots + b_1 X + b_0$. (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 $\alpha^n = -(b_{n-1} \alpha^{n-1} + \cdots + b_1 \alpha + b_0)$. If the product g(α)h(α) has a term αm with mn it can be reduced as follows:

$\alpha^n\alpha^{m-n} = -(b_{n-1} \alpha^{n-1} + \cdots + b_1 \alpha + b_0) \alpha^{m-n} = -(b_{n-1} \alpha^{m-1} + \cdots + b_1 \alpha^{m-n+1} + b_0 \alpha^{m-n+1})$.

As an example of the reduction rule, take $K_i = \mathbb Q[X]$, the ring of polynomials with rational coefficients, and take f(X) = X7 − 2. Let g(α) = α5 + α2 and h(α) = α3 + 1 be two elements of $\mathbb Q[X]/(X^7-2)$. The reduction rule given by f(X) is α7 = 2 so

$g(\alpha) h(\alpha) = \left(\alpha^5 + \alpha^2\right) \left(\alpha^3 + 1\right) = \alpha^8 + 2 \alpha^5 + \alpha^2 = \left(\alpha^7\right) \alpha + 2\alpha^5 + \alpha^2 = 2 \alpha^5 + \alpha^2 + 2\alpha.$

## Examples

### The complex numbers

Consider the polynomial ring R[x], and the irreducible polynomial x2 + 1. The quotient space 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, x5x, etc.; and so, for example p + qx + rx2 + sx3p + qx + r⋅(−1) + s⋅(−x) = (pr) + (qs)⋅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, x5x, etc. Thus:

$(a_1 + b_1x) + (a_2 + b_2x) = (a_1 + a_2) + (b_1 + b_2)x \, ,$
$(a_1 + b_1x)(a_2 + b_2x) = a_1a_2 + (a_1b_2 + b_1a_2)x + (b_1b_2)x^2 \equiv (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)x \, .$

If we identify a + bx with (a,b) then we see that addition and multiplication are given by

$(a_1,b_1) + (a_2,b_2) = (a_1 + a_2,b_1 + b_2) \, ,$
$(a_1,b_1)\cdot (a_2,b_2) = (a_1a_2 - b_1b_2,a_1b_2 + b_1a_2) \, .$

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 + ib, where a and b are real numbers and i2 = −1. Addition and multiplication are given by

$(a_1 + ib_1) + (a_2 + ib_2) = (a_1 + a_2) + i(b_1 + b_2) \, ,$
$(a_1 + ib_1) \cdot (a_2 + ib_2) = (a_1a_2 - b_1b_2) + i(a_1b_2 + a_2b_1) \, .$

If we identify a + ib with (a,b) then we see that addition and multiplication are given by

$(a_1,b_1) + (a_2,b_2) = (a_1 + a_2,b_1 + b_2) \, ,$
$(a_1,b_1)\cdot (a_2,b_2) = (a_1a_2 - b_1b_2,a_1b_2 + b_1a_2) \, .$

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 + bxa + ib is a homomorphism with respect to addition and multiplication. It is also obvious that the map a + bxa + ib is both injective and surjective; meaning that a + bxa + ib is a bijective homomorphism, i.e. an isomorphism. It follows that, as claimed: R[x] / (x2 + 1) ≅ C.

### Cubic example

If K is the rational number field Q and

p(X) = X3 − 2,

then a splitting field L will contain a primitive cube root of unity,[why?] as well as a cube root of 2. Thus

${L=\mathbb{Q}(\sqrt[3]{2},\omega_2)=\{a+b \omega_2+c\sqrt[3]{2} +d \sqrt[3]{2} \omega_2+ e \sqrt[3]{2^2} + f \sqrt[3]{2^2} \omega_2 \,|\,a,b,c,d,e,f\in\mathbb{Q} \}}$

where

$\omega_1 = 1 \,$,
$\omega_2 = - \frac{1} {2} + \frac {\sqrt{3}} {2} i$, and
$\omega_3 = - \frac{1} {2} - \frac {\sqrt{3}} {2} i$

are the cubic roots of unity.

### Other examples

• A splitting field of x2 + 1 over $\mathbb F_7$ is $\mathbb F_{49}$; the polynomial has no roots in $\mathbb F_7$, i.e., −1 is not a square there, because 7 is not equivalent to 1 (mod 4).[1]
• The splitting field of x2 − 1 over $\mathbb F_7$ is $\mathbb F_7$ since x2 − 1 = (x + 1)(x − 1) already factors into linear factors.

## References

1. ^ Instead of applying this characterization of odd prime moduli for which −1 is a square, one could just check that the set of squares in $\mathbb F_7$ is the set of classes of 0, 1, 4, and 2, which does not include the class of −1≡6.
• Dummit, David S., and Foote, Richard M. (1999). Abstract Algebra (2nd ed.). New York: John Wiley & Sons, Inc. ISBN 0-471-36857-1.

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