Fields and Field Extensions

Field Extensions and Common Fields

Field Extension

The field is an extension of the field if . We write to denote this extension.

Here are the three most important classes of fields:

Number Fields, Finite Fields and Function Fields

A number field is a subfield of . A finite field is a field with finitely many elements. A function field is an extension field of of rational functions in one variable.

Algebraic and Transcendental Elements

Algebraic & Transcendental

Suppose is an extension of a field . An element is algebraic over if there exists a nonzero monic polynomial such that . Otherwise, it is called transcendental.

e.g. is algebraic over but transcendental over .

Proposition

Suppose is an extension of a field . Fix some , consider Then is algebraic over if and only if . Equivalently, is transcendental over if and only if is injective.

Proof This is a direct consequence of the definition.

Proposition

Fix some algebraic over . Then the following are equivalent for a monic polynomial :

• is the monic polynomial of lowest degree such that .
• is irreducible in and .
• , and is a maximal ideal in .
• , and whenever for some , then .

Proof The equivalence of the first two statements, and the last two statements are straightforward. We will prove

Minimal Polynomial & Degree of an Algebraic Element

The unique monic polynomial in the above proposition is called the minimal polynomial of over , and its degree is called the degree of over , denoted by .

Generated Field Extension

Let be an extension field of . The subfield of generated by and an element is denoted by . That is, is the smallest subfield of containing and . Similarly, is the smallest subfield containing and .

Remark

Note that we denote as the smallest ring containing and . It is not necessarily a field. Indeed, .

Proposition

Let be an element of an extension field which is algebraic over , and let be the irreducible polynomial for over . Then

• The canonical map is an isomorphism, and is a field. Thus .
• More generally, let be elements of an extension field , which are algebraic over . The ring is equal to the field .

Proof Let be the map, as is irreducible, the ideal is maximal, and so . Then by the first isomorphism theorem, , is a field. The second statement follows from induction: $$F[\alpha_1,\ldots,\alpha_k]=F[\alpha_1,\ldots,\alpha_{k-1}][\alpha_k]=F(\alpha_1,\ldots,\alpha_{k-1})\left[\alpha_k\right]=F(\alpha_1,\ldots,\alpha_n).$$$\square$

Proposition

Let be an algebraic element over , and let be the irreducible polynomial for over . If has degree , i.e., if has degree over , then is a basis for as a vector space over .

Proof Firstly, it is obvious to see that is a vector space over , particularly, it is spanned by . Suppose there is a linear dependence relation where not all are zero, then it contradicts the minimality of by division with remainder.

e.g. Suppose . Then is a basis for over .

Remark

It may not be easy to tell whether two algebraic elements and generate isomorphic field extensions, but the above proposition provides a necessary condition: they must have the same degree over .

Isomorphism of Field Extensions

Let and be extensions of the same field . An isomorphism that restricts to the identity on the subfield is called an -isomorphism, or an isomorphism of field extensions. If there exists an -isomorphism , and are isomorphic extension fields.

Proposition

Let be a field, and let and be elements of field extensions and . Suppose that and are algebraic over . There is an -isomorphism of fields that sends if and only if the irreducible polynomials for and over are equal.

Proof Let the shared irreducible polynomial be , then under the canonical isomorphisms.

Proposition

Let be an -isomorphism, and let . Let be a root of in , and let be its image in . Then is also a root of .

Proof Suppose , then $$f(\alpha)=\sum_{i=0}^{n}a_{i}\alpha^{i}=0\implies \sum_{i=0}^{n}a_{i}\varphi(\alpha)^{i}=0.$$$\square$

Degrees of Field Extensions

Now we generalize the idea of the proposition to any field extension. Note that a field extension of can always be regarded as an -vector space. Addition is the addition law in , and scalar multiplication of an element of by an element of is obtained by multiplying these two elements in .

Degree of a Field Extension

 Suppose is a field extension. Then the dimension of , when regarded as an -vector space, is called the degree of the field extension. This degree is denoted by or .  A field extension is a finite extension if its degree is finite. Extensions of degree are quadratic extensions, those of degree are cubic extensions, and so on.

Lemma

The followings are true:

• A field extension has degree if and only if .
• An element of a field extension has degree over if and only if is an element of .

Proof If , then the set is a basis of over , so . If has degree over , then its minimal polynomial must be , so , and vice versa.

Proposition

are fields. Suppose is algebraic over . Then if and only if the minimal polynomial of is of degree .

Proof Let be the minimal polynomial of over . Then , so .

Characteristic

Let be the unique ring homomorphism from the integers to a ring . Then the characteristic of is the smallest nonnegative integer such that . i.e. is the smallest positive integer such that in , or if no such integer exists.

Proposition

Assume that the field does not have characteristic , that is, in . Then any extension of degree over can be obtained by adjoining a square root: , where is an element of . Conversely, if is an element of a field extension of , and if is in but is not in , then is a quadratic extension of .

Remark

However, it is not true that all cubic extensions can be obtained by adjoining a cube root.

Theorem

If is a field extension of , and , then every element of is algebraic over .

Proof Assume there is a transcendental element . Then there holds Note that has infinite degree, so , which is a contradiction.

Multiplicative Property of the Degree

Suppose , where , are fields and is a ring. Then

Proof Suppose and are bases of over and over respectively. Then is a basis of over .

One takeaway from the above theorem is that has to divide . So we have the following corollaries:**

Corollary

Let be an extension field of of prime degree . If an element of is not in , then has degree over and .

Corollary

Let be a sequence of field extensions. If an element is algebraic over , then is algebraic over and is at most .

Proof Let be the minimal polynomial of over . Since , is also a polynomial over . Since is a root of , the minimal polynomial of over divides , so .

Corollary

Let and be finite extensions of , and let denote the field generated by the two fields and together. Let , and . Then and divide , and .

Proof The multiplicative property shows that and divides . Now suppose . Then the corollary shows that , similarly , so . The general case follows by induction.

So one can apply the above proposition to help to find the minimal polynomial of an algebraic element over a field.

e.g.

Ruler and Compass Constructions

Ruler and Compass Constructions

A ruler and compass construction is a method of drawing geometric figures using only a straightedge (ruler) and a compass.

• Two points in the plane are given to start the construction.
• The straightedge can be used to draw a line between any two points.
• The compass can be used to draw a circle with a given center and a point on the circle.

We say the points of intersection of the lines and circles are constructed points. Points, lines, and circles will be called constructible if they can be obtained in finitely many steps, using these rules.

Theorem

Let be a constructible point in the plane. Then there is a chain of field extensions such that the coordinates of are in , and each extension is a quadratic extension.

Corollary

Let be a constructible real number. Then is algebraic over and the degree of over is a power of .