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"Professor" Andrey Mishchenko Math 296 Lecture Notes

Nick Wasylyshyn January 24-28, 2010

January 25 : Complex Preliminary notes There will be a quiz tomorrow ! Review notes and definitions and it will be easy.

What is a over C ? Definition : f is a polynomial over C if f : C → C is given by n n−1 f(z) = anz + an−1z + ··· + a0 for some fixed a0, a1, . . . , an ∈ C.

Two Key Theorems from Last Class Theorem 1 : Every non-constant polynomial over C has a root over C. Theorem 2 : Given f, a polynomial over C, there is another polynomial g over C such that f(z) = g(z)(z − r) if and only if r ∈ C is a root of f. Theorem 1 is known as the Fundamental Theorem of Algebra, which can be more pre- cisely stated as follows : The Fundamental Theorem of Algebra : Every polynomial over C of degree n ∈ N has n roots in C, counting multiplicity. A trivial example of Theorem 2 is as follows : Example : Let f be the polynomial defined by f(x) = x2 − 3x − 4. Since x = 4 is a root of this polynomial, we can find a polynomial g, namely g(x) = x + 1, such that (x − 4)g(x) = f(x).

An Important Consequence of Theorem 2 Any polynomial over C uniquely splits as a product of linear terms. Note : The uniqueness of this split will be discussed later. To understand this fact, it is useful to know precisely what it means for a polynomial to split completely over a field. So, precisely and more generally, Definition : A polynomial splits completely over a field F if f(x) can be written as c(x − r1)(x − r2) ··· (x − rn), where c, r1, r2, . . . , rn ∈ F. This means that for any polynomial f over C, f(x) can be written as c(x − r1)(x − r2) ··· (x − rn), where c, r1, r2, . . . , rn ∈ C.

1 Question : Do polynomials over R split over R ? Sadly, no. For instance, the only way to split x2 +1 into linear terms is x2 +1 = (x+i)(x−i). As i∈ / R, x2 + 1 does not split over R. DJ asked how often polynomials over R split over R. This question is for a different time, but the answer to the related question "Can we easily tell whether a polynomial splits over R ?" is no.

Field To understand the next section, we must digress a bit to define the terms field and field . Definition : Two fields F1 and F2 are isomorphic with the isomorphism f : F1 → F2 if the following conditions are met :

1. f(0F1 ) = 0F2

2. f(1F1 ) = 1F2 3. f(a + b) = f(a) + f(b) 4. f(ab) = f(a)f(b) Definition : A field automorphism is a field isomorphism from one field to itself. So, a field automorphism of C is a field isomorphism f : C → C.

Complex Conjugation Definition : For z = a + bi ∈ C, the complex conjugate of z, denoted z¯ or z∗, equals a − bi. Having defined field , we can now discuss the following theorem : Theorem 3 : The function f : C → C defined by f(z) =z ¯ is a field automorphism of C. Proof : We must test the four conditions. Conditions 1 and 2 are trivial. Condition 3 is easily proved as follows : Pick z = a + bi, w = c + di ∈ C. f(z + w) = z + w = a + bi + c + di = (a + c) + (b + d)i = (a + c) − (b + d)i = (a − bi) + (c − di) =z ¯ +w ¯. The proof of condition 4 is also simple, and is left to the reader as an exercise√ similar to a part of problem 3 of homework 1 of Math 295, which dealt with the field Q( 2).

Splitting Polynomials into Quadratic Terms We know from Theorems 1 and 2 that every polynomial over R of degree ≥ 1 has a linear factor in C. It is also possible to show that every polynomial over R of degree ≥ 2 has a real quadratic factor. n n−1 Theorem 4 : If f is a polynomial over R defined by f(x) = anx + an−1x + ··· + a0, where a0, . . . , an ∈ R and n ≥ 2, then f has a real quadratic factor. Proof : Let z = a + bi ∈ C be a root of f. Note : by the Fundamental Theorem of Algebra, z exists. n n−1 Since z is a root of f, we know that f(z) = anz + an−1z + ··· + a0 = 0. Let’s take the conjugate of both sides. n n−1 0¯ = 0 = anz¯ + an−1z¯ + ··· + a0 Note : We can break up the conjugation of the right side of the equality as we did because we know that complex conjugation is a field automorphism of C, so it distributes over addition and multiplication.

2 Since all a0, . . . , an ∈ R, ∀i ∈ 0, 1, . . . , n, ai = ai, so we can rewrite the write side of our n n−1 n n−1 equality as anz¯ + an−1z¯ + ··· + a0. Hence, anz¯ + an−1z¯ + ··· + a0 = f(¯z) = 0, so z¯ is a root of f, too ! So we can factor out (x − z)(x − z¯) from f(x). We want to show that (x − z)(x − z¯) ∈ R. (x − z)(x − z¯) = x2 − zx¯ − zx + zz¯ Note : earlier we defined z = a + bi, so z¯ = a − bi. – Clearly, x2 ∈ R. – −zx¯ − zx = −x(¯z + z) = −x(a − bi + a + bi) = −x(2a) ∈ R, since a ∈ R. – zz¯ = (a + bi)(a − bi) = a2 + b2 ∈ R. So, since (x−z) and (x−z¯) are factors of f and (x−z)(x−z¯) ∈ R, f has the real quadratic factor (x − z)(x − z¯).

A Few Interesting Tidbits Question : We essentially created C by taking R and throwing in i and adding it to and multiplying it by itself and by elements of R. What if instead of throwing in i, we added a root of a different polynomial ? Could we get back to i to construct C ? √ √ For example, what about −17 ? This works, because √−17 = ±i, and from this we can 17 get the rest of C. What about the generic root r of some polynomial f over R ? According to the Fundamental Theorem of Algebra, r ∈ C, which means that for some a, b ∈ R, r = a + bi. Thus, we can simply subtract a and divide by b to get back to i, and from there we can construct the rest of C.

Finding Roots of Polynomials

We all learned the quadratic formula in middle school, which shows us explicitly the roots of a polynomial of degree 2. There is an analogous "cubic formula" and "quartic formula", although they are much longer and more difficult to sing than the quadratic formula. However, Theorem 5 : There is no formula that can be used to find solutions, expressible by radicals, of polynomials whose degree is at least 5. The math needed to prove this theorem would take a semester to learn, but if the theorem intrigues you, Prof. Andrey suggests you check out the book Galois Theory by Ian Stewart, which does a good job of walking the reader through math like this.

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