Algebraic Extensions
In this first chapter concerning polynomial equations, we show that given a polynomial over a field, there always exists some extension of the field where the polynomial has a root, and we prove the existence of an algebraic closure. We make a preliminary
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Algebraic Extensions
In this first chapter concerning polynomial equations, we show that given a polynomial over a field, there always exists some extension of the field where the polynomial has a root, and we prove the existence of an algebraic closure. We make a preliminary study of such extensions, including the automorphisms, and we give algebraic extensions of finite fields as examples.
§1.
FINITE AND ALGEBRAIC EXTENSIONS
Let F be a field. If F is a subfield of a field E, then we also say that E is an extension field of F . We may view E as a vector space over F , and we say that E is a finite or infinite extension of F according as the dimension of this vector space is finite or infinite. Let F be a subfield of a field E. An element (I( of E is said to be algebraic over F if there exist elements ao, ... , an (n ~ 1) of F, not all equal to 0, such that
If (I( ¥- 0, and (I( is algebraic, then we can always find elements a, as above such that ao ¥- (factoring out a suitable power of (I(). Let X be a variable over F. We can also say that (I( is algebraic over F if the homomorphism F[X] -. E
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223 S. Lang, Algebra © Springer Science+Business Media LLC 2002
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ALGEBRAIC EXTENSIONS
which is the identity on F and maps X on ex has a non-zero kernel. In that case the kernel is an ideal which is principal, generated by a single polynomial p(X), which we may assume has leading coefficient 1. We then have an isomorphism F[X]/(p(X)) ~ F[ex], and since F[ex] is entire, it follows that p(X) is irreducible. Having normalized p(X) so that its leading coefficient is 1, we see that p(X) is uniquely determined by ex and will be called THE irreducible polynomial of ex over F. We sometimes denote it by Irr(ex, F, X). An extension E of F is said to be algebraic if every element of E is algebraic over F.
Proposition 1.1. Let E be a finite extension of F. Then E is algebraic over F. Proof.
Let ex E E, ex ¥= 0. The powers of ex,
cannot be linearly independent over F for all positive integers n, otherwise the dimension of E over F would be infinite. A linear relation between these powers shows that ex is algebraic over F. Note that the converse of Proposition 1.1 is not true; there exist infinite algebraic extensions . We shall see later that the subfield of the complex numbers consisting of all algebraic numbers over Q is an infinite extension ofQ. If E is an extension of F, we denote by [E:F]
the dimension of E as vector space over F . It may be infinite.
Proposition 1.2. L et k be a field and FeE extension fields of k. Then
[E : k] = [E : F] [F :k]. If
is a basis for F over k and {XiYj}(i .j)el xJ is a basis for E over k. {X i} iel
{Yj}jeJ
is a basis for E over F, then
Proof. Let Z E E. By hypothesis there exist elements exj E F, almost all exj = 0, such that
For each j such that
E
J there exist elements bji E k, almost all of which are equal to 0,
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FINITE AND ALGEBRAIC EXTENSIONS
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and hence
This shows that {XiYj} is a family of generators for E over k. We must show that it is linearly
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