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Polynomial: Conjugate Root Theorem
Motivation: Have you ever noticed that : if you have one root for a polynomial, for example, say the the root is (2- ), then (2+ ) will also be a root of that equation. Or sometimes (2-3i) just appears (2+3i) . Are there any rules or constraints behind that? What is the situation that this will happen? Let’s check the following two cases:
1. x2 – 5x + 3 = 0 We have x = So, ½( 5+ ) and ½(5- ) appears together in this case.
2. x2 - x + 1 = 0 In this case, x =
Notice the difference ? When you see one root is ½( +1), the other root is NOT ½(1- ); it is ½( -1) .
Definition ( Irreducible Polynomial , Revisited ) f(x) F[x]. If f(x) has no non-constant factor with smaller degree in F[x], we say that f(x) is irreducible in F[x].
Recall that : when we say a polynomial is irreducible, we need to specify the “scope” very clearly. An example is (x2 -2 ) is irreducible in Q[x], but it is not irreducible in R[x] : (x2-2) = ( x + ) ( x- )
It goes without saying that Z Q R C. ( The notation “A B” means “B contains A” )
When a polynomial is irreducible in a smaller scope, it probably would not be irreducible in a larger scope. But that is also why the following theorem is valuable.
Theorem: If a polynomial f(x) with integer coefficients is irreducible in Z[x], then it is also irreducible in Q[x].
Proof: We just state the sketch without details. The complete proof is a little lengthy. Basically, it uses “Gauss Lemma” to prove this theorem.
Consider the case if f(x) is irreducible in Q[x], of course it is impossible to find several lower order of polynomials in Z[x] such that their product is f(x). So, we consider the cases that f(x) can be represented into the product of several non-trivial lower order polynomials in Q[x]. For those non-trivial lower order polynomials, we can convert those polynomials into polynomials with integer coefficients by multiplying the lcm of denominators and divided the gcd of numerators. Finally, it can be organized into the following f(x) = m p(x)q(x) where p(x), q(x) Z[x] gcd of coefficients of p(x) = 1 gcd of coefficients of q(x) = 1 m Q
And we hope that we can claim that m is an integer. For rigorous proof, you can look into “Gauss Lemma” from elsewhere. Intuitively, you can think the equality will hold for any x. Because f(x) is with integer coefficients, p(x) and q(x) are also with integer coefficients, if m is not an integer, the equality might not hold when we set x to any integer – the left hand side of the equality will always be an integer for any integer x.
Definition ( Minimum Polynomial ) Let L K and c K, but c L. The minimum polynomial of c in L is defined as the polynomial with the lowest order such that the coefficients of the polynomial are in L and c is a root of that polynomial.
Example: the minimum polynomial of in Q is (x2 -2 ) . ( We know that is not in Q )
Theorem: Minimum polynomial is irreducible in L. Proof: Assume that a minimum polynomial m(x) is not irreducible in L, i.e., there exist p(x), q(x) in L such that
m(x) = p(x)q(x) where p(x), q(x) L[x] and p(x) and q(x) are not trivial
But m(c) = 0 p(c) q(c) = 0 p(c) = 0 or q(c) = 0
And we know that it is impossible because m(x) is with the lowest order such that m(c) = 0. the assumption does not hold.
So, m(x) is irreducible in L.
Definition ( Conjugate Elements ) The conjugate elements of c in K are the other roots of the minimum polynomial in L.
Theorem ( Conjugate Root Theorem ): Let L K and c K, but c L . For any polynomial f(x) in L[x], if f(c) = 0, then the other conjugate elements of c are also the roots of f(x).
Proof: The minimum polynomial ( denoted as m(x) ) of c in L is irreducible in L[x]. m(x) | f(x) if f(c) = 0.
So, the other conjugate elements of c are also the roots of f(x).
Example: Find the minimum polynomial of (c + ) in Q[x], where d > 0 and d is not a square of any rational number and c, d Q.
Sol: x = c+ x-c = To eliminate the square root sign, there is no choice but squaring both sides:
(x-c)2 = d x2 – 2cx + (c2-d) = 0 And this equation is with roots (c+ ) and (c- ) .
So, if any polynomial in Q[x] containing the root (c+ ), it must also contains (c- ) .
However, for polynomial in R[x], it is not necessary to be true. You can create a polynomial like (x-c- )(x-c+5 ); it does not have to contain ( c- ).
Note: this example only tells that for any polynomial in Q[x] with root (c+ ), it must have root (c- ). But it does not tell you all the roots should be of this form. Please do not be confused. You could have roots in other form. For example, is a root of
x3 – 2 = 0 .
Example: Find the minimum polynomial of (c+di) in R[x], where d 0, c, d R, i = Sol: x = c + di x-c = di To eliminate i , there is no choice but taking square both sides: (x-c)2 = -d2 (x-c)2 + d2 = 0 x2 – 2cx + (c2+d2 ) = 0 it is with roots (c+di) and (c-di) .
So, for any polynomial in R[x] with root (c+di), it must also contain (c-di) . But for polynomial in C[x], it is not the case. You can have a polynomial like (x-c-di)( x-c-2di) without containing (c-di) as a root.
Exercise: Find the minimum polynomial of ( + ci) in Q[x], where c,d Q, d > 0, d is not a square of any rational number, and i = .
And from that, you can conclude that for any polynomial in Q[x] containing the root ( +ci), what will be the other roots in that polynomial?
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