Minimal Polynomials

[Definitions & Properties] [Exensions] [Which real numbers are constructible?]

Before reading this section, you should have an understanding of Rings and Fields.

Definitions & Properties

F[x]

Where F is a field, F[x] is the ring of polynomials in x over F.

Degree of a Polynomial

If p(x) = a_0 + (a_1)x + ... + (a_n)x^n and a_n =/ 0, then the degree of p(x), deg p(x), is n. (deg(x^3 + 3x^2 + 1) = 3)

Monic Polynomial

f(x) is in F[x] is a monic polynomial if the coefficient of its highest power is 1.

Minimal Polynomial

A polynomial p(x) in F[x] of degree n, is a minimal polynomial if p(a) = 0, no nonzero polynomial, q(x), of lower degree in F[x] has the property that q(a) = 0 and p(x) is monic.

Principal Ideal Domain

An integral domain R is a principal ideal domain if every ideal I in R is of the form I = {xa| x R} for some a in I.

Irreducible

The polynomial p(x) in F[x] is irreducible if

(1) p(x) is of positive degree

(2) given any polynomial f(x) in F[x], either p(x)|f(x) or p(x) is relatively prime to f(x)

Plane of B

For any subfield B R, let P(B), the plane of B, = B x B R^2.

Properties of Polynomial Rings

Proof: The set of polynomials forms a ring in x over F

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Extensions

Algebraic

If K F are fields, then a K is said to be algebraic over F if there exists a polynomial p(x) 0 in F[x] such that p(a) = 0.

Algebraic Degree

The element a in the extension K of F is said to be algebraic of degree n if there is a polynomial p(x) in F[x] of degree n such that p(a) = 0 and no nonzero polynomial of lower degree in F[x] has this property.

Extension

F(a) is called the field or extension obtained by adjoining a to F. F is an extension of K if F K where the operations of F are those restricted to elements in K.

Finite Extensions

F is a finite extension of K if the degree of F over K is finite.

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Which real numbers are constructible?

We now know that the constructible numbers form a subfield of the field of real numbers. The question now becomes whether or not the constructible numbers are themselves the field of real numbers.

As you might have guessed from the introduction, we can't construct all of the real numbers but why not?

Let's first use a straight-edge to draw a segment. Let's label this segment as length 1. We can copy this segment and then add the two segments together. If e continually add a segment of length 1, we can construct the natural numbers. Instead of adding two segments, we can subtract two segments, this is how we form all of the integers.

Since we can construct the integers and we know we can construct a/b where a and b are integers, we can construct the rational numbers. So what it really comes down to is whether or not we can construct all of the irrational numbers.

If we create 1 then we can double that length and construct 2. If we construct a perpendicular bisector of the line then we have created a 90 degree angle. Using our compass we can construct length one on the perpendicular line starting at the intersection of the two lines. Now using a ruler we can connect the endpoints of two two lengths and we have created a 45-45-90 triangle. Two of the sides are of length one so the third side will be sqrt(2). Constructing a 45-45-90 triangle.

Since we can construct sqrt(2), does this mean we can construct all of the irrational numbers and thus all of the real numbers?

It turns out that in order for a real number to be constructible the minimal polynomial of a over the rationals must have degree a power of 2.

Looking at the number sqrt(3) + sqrt(2). The degree of the minimal polynomial over the rationals will be a^2 = 5 + 2sqrt(6), (a^2 - 5)^2 = 12, a^4 - 10a^2 +13 = 0. So the minimal polynomial has degree 4 so sqrt(3) + sqrt(2) is constructible.

How about cube root of 3? a^3 - 3 will be the minimal polynomial so the degree of the minimal polynomial over the rationals will be 3 but 3 is not a power of 2. Therefore the cube root of 3 is not a constructible number.

Here's a proof:

Theorem: In order for the real number a be constructible, it is necessary that [Q(a):Q] be a power of 2 where Q is the set of rationals.

Proof:

Let C be the set of all constructible numbers. Let C_1 be a subfield of C. Let P(C_1) = C_1 x C_1, be a subset of C^2, such that every point in P(C_1) can also be represented as (a, b) where a and b are in C_1. We must show that if we draw two figures, given a straight-edge and a compass, their points of intersection will be in P(C_1). The only two shapes we can draw are straight lines and circles, the straight-edge allows us to draw straight lines and the compass lets us draw circles. So there are three possible results:
1.Given two straight lines, both intersecting two points in P(C_1), the intersection point, if any, of the two lines will be in P(C_1).
2. Given a straight line that intersects two points in P(C_1) and a circle whose radius is in C_1, their intersection point(s) will be in P(C_1) or P(C_2) where [C_2|C_1] = 2.
3. Given two circles whose radii are both in C_1, their intersection point(s) will be in P(C_1) or P(C_2) where [C_2|C_1] = 2.

Given two straight lines, y = Ux + V and y = Wx + Z, where U, V, W, and Z are in C_1, they will intersect at most once. If U = W then the two lines will run parallel and never intersect, otherwise these two lines will intersect exactly once. Subtracting the two equation establishes x = (Z – V)/(U-W), so x will be in C_1. Plugging x back into one of the original equations will give us y in C_1. So the point of intersection (x, y) will be in P(C_1).

Next we must look at the circumstance where a circle intersects a line. Let the radius of the circle be r and center be (a, b). So the equation can be written as (x – a)^2 + (y – b)^2 = r^2. The equation of the line is of the form y = Ux + V where U and V are in C_1. Plugging y into the equation of the circle will give us (x – a)^2 + (Ux + V – b)^2 = r^2, x^2 – 2ax + a^2 + (U^2)x^2 + 2Uvx – 2Ubx – 2bV + b^2 = r^2, (1 + (U^2))x^2 + (2Uv – 2a – 2Ub)x + (a^2 – 2bV + b^2 – r^2) = 0 which is a quadratic equation of the form w^2 + c_1(w) + c_2 where w is the x-coordinate of the intersection point and c_1 and c_2 are in C_1. By the quadratic formula we know that if the line and circle are going to intersect in P(C_2), sqrt(((c_1)^2 – 4(c_2))/2) ≥ 0. Since c_1 and c_2 are in C_1, if C_2 = C_1(sqrt(c)) and if sqrt(((c_1)^2 – 4(c_2))/2) ≥ 0, then w is in C_1 and [C_2:C_1] = 1. Another option is for C_1 = C_2 if sqrt(C_1) is in C_1 so [C_2| C_1] = 1. Otherwise the extension field of [C_2:C_1] = 2. Now that we know that x-coordinate is in C_1 we must make sure that the y-coordinate is in C_1. Since z = Uw + V, where U, w, and V are in C_1, z is in C_1. So now we know that (w, z) will be in P(C_1) or P(C_2) and [C_2: C_1] = 1 or 2.

Our last possibility is for two circles to intersect. If x^2 + y^2 + dx + ey + f = 0 and x^2 + y^2 + gx + hy + j = 0 then by subtracting these two equations we get the equation of a line, (d-g)x + (e-h)y + (f – h) = 0. We can now revert back to the case where a line is intersecting a circle. So the point(s) of intersection of the two circles will be in P(C_1) or P(C_2) and [C_2:C_1] = 1 or 2.

The question that we started with was can we construct all of the real numbers. So let’s construct a length c. Using a straight-edge we can construct lines of length a/b where a and b are integers, b 0. This gives us the set of rational numbers. On a similar note we can use the compass to construct circles in the plane of rational numbers. We know from above that a circle and line in the plane of rational numbers will intersect at a point in the plane whose extension field is of degree 1 or 2. Let’s assume that z, the length we want to construct, is in the plane, L_n. In order to get to the plane L_n from the plane of rationals, we must first go to L_1. The extension of L_1 over the rationals will have degree 1 or 2. From L_1 we must go to L_2. The extension of L_2 over L_1 has degree 1 or 2. We can continue this process a finite number of times until we eventually get to the plane L_n where L_0 L_1 L_2 L_3 ... L_ n equals the rationals. So the extension of L_n over the rationals is a power of 2 since the extension of L_n over the rationals is equal to the extension of L_1 over the rationals times the extension of L_2 over L_1 times … times the extension of L_(n) over L_(n-1) where each of these extension fields are of degree 1 or 2. Since z is in L_n, the extension adjoining z to the rationals will be a subfield of L_n and the minimal polynomial of z over the rationals must have a degree z power of 2.

This proof was based on the proof given on page 204 in the Third Edition of Abstract Algebra by I.N. Herstein.

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