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Splitting Fields

Let F be a field and p(x) be a nonconstant polynomial in F[x]. We already know that we can find a field extension of F that contains a root of p(x).

Splitting Fields

Let \(F\) be a field and \(p(x)\) be a nonconstant polynomial in \(F[x]\). We already know that we can find a field extension of \(F\) that contains a root of \(p(x)\). However, we would like to know whether an extension \(E\) of \(F\) containing all of the roots of \(p(x)\) exists. In other words, can we find a field extension of \(F\) such that \(p(x)\) factors into a product of linear polynomials? What is the smallest extension containing all the roots of \(p(x)\)?

Let \(F\) be a field and \(p(x) = a_0 + a_1 x + \cdots + a_n x^n\) be a nonconstant polynomial in \(F[x]\). An extension field \(E\) of \(F\) is a splitting field of \(p(x)\) if there exist elements \(\alpha_1, \ldots, \alpha_n\) in \(E\) such that \(E = F( \alpha_1, \ldots, \alpha_n )\) and \[\begin{aligned}\end{aligned}\]. A polynomial \(p(x) \in F[x]\) splits in \(E\) if it is the product of linear factors in \(E[x]\).

Example

Let \(p(x) = x^4 + 2x^2 - 8\) be in \({\mathbb Q}[x]\). Then \(p(x)\) has irreducible factors \(x^2 -2\) and \(x^2 + 4\). Therefore, the field \({\mathbb Q}( \sqrt{2}, i )\) is a splitting field for \(p(x)\).

Example

Let \(p(x) = x^3 - 3\) be in \({\mathbb Q}[x]\). Then \(p(x)\) has a root in the field \({\mathbb Q}( \sqrt[3]{3}\, )\). However, this field is not a splitting field for \(p(x)\) since the complex cube roots of \(3\), \[\begin{aligned}\end{aligned}\], are not in \({\mathbb Q}( \sqrt[3]{3}\, )\).

Theorem

Let \(p(x) \in F[x]\) be a nonconstant polynomial. Then there exists a splitting field \(E\) for \(p(x)\).

Proof

We will use mathematical induction on the degree of \(p(x)\). If \(\deg p(x) = 1\), then \(p(x)\) is a linear polynomial and \(E = F\). Assume that the theorem is true for all polynomials of degree \(k\) with \(1 \leq k \lt n\) and let \(\deg p(x) = n\). We can assume that \(p(x)\) is irreducible; otherwise, by our induction hypothesis, we are done. By , there exists a field \(K\) such that \(p(x)\) has a zero \(\alpha_1\) in \(K\). Hence, \(p(x) = (x - \alpha_1)q(x)\), where \(q(x) \in K[x]\). Since \(\deg q(x) = n -1\), there exists a splitting field \(E \supset K\) of \(q(x)\) that contains the zeros \(\alpha_2, \ldots, \alpha_n\) of \(p(x)\) by our induction hypothesis. Consequently, \[\begin{aligned}\end{aligned}\] is a splitting field of \(p(x)\).

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

imaginary unit
i² = −1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

a \mid b,\ \gcd(a,b)

divides, greatest common divisor
b is a multiple of a; the largest number dividing both.

(G, \cdot),\ e,\ g^{-1}

group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.

G \cong H,\ G / N

isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.

\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n

integers modulo n
The remainders 0…n−1 with clock arithmetic.

\operatorname{Hom}(A, B),\ f \circ g

arrows from A to B, composition
The set of morphisms; do g then f.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Özüňi synla

Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

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