Then p � 0 so there is a monic polynomial in K z with � as a root Pick a monic

# Then p ? 0 so there is a monic polynomial in k z with

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Then p ( λ ) = 0, so there is a monic polynomial in K [ z ] with λ as a root. Pick a monic polynomial q K [ z ] of minimal degree such that q ( λ ) = 0. Then obviously q is irreducible. Uniqueness: Suppose p, q K [ z ] are both monic polynomials of minimal degree such that p ( λ ) = q ( λ ) = 0. Then deg( p ) = deg( q ), so by the Euclidean algorithm, there are k, r K [ z ] with deg( r ) < deg( q ) such that p = kq + r . Now p ( λ ) = k ( λ ) q ( λ ) + r ( λ ), so r ( λ ) = 0. This is only possible if r = 0 as deg( r ) < deg( q ). Hence p = kq with deg( p ) = deg( q ), so k is a constant. As p, q are both monic, k = 1, and p = q . Corollary 4.3.3. Suppose λ F . Then λ K if and only if deg(Irr K ) = 1 . Corollary 4.3.4. Suppose λ F and dim K ( F ) = n . Then deg(Irr K ) n . Lemma 4.3.5. Suppose λ F is a root of p K [ x ] . Then Irr K | p . Proof. The proof is similar to 4.3.2. Without loss of generality, we may assume p is monic (if not, we replace p with p/ LT( p ). Now since deg(Irr K ) deg( p ), by the Euclidean algorithm, there are k, r K [ z ] with deg( r ) < deg(Irr K ) such that p = k Irr K + r . We immediately see r ( λ ) = 0, so r = 0. Example 4.3.6. We know that C is a two dimensional real vector space. If λ C is a root of p R [ z ], then λ is also a root of p . In fact, since p ( λ ) = 0, we have that p ( λ ) = p ( λ ) = 0 since the coefficients of p are all real numbers. Hence, the irreducible polynomial in R [ z ] for λ C is given by Irr R ( z ) = ( z - λ if λ R ( z - λ )( z - λ ) = z 2 - 2 Re( λ ) z + | λ | 2 if λ / R . Definition 4.3.7. The polynomial p F [ z ] splits (into linear factors) if there are λ 1 , . . . , λ n F such that p ( z ) = n Y i =1 ( z - λ i ) . Examples 4.3.8. (1) Every constant polynomial trivially splits in F [ z ]. (2) The polynomial z 2 + 1 splits in C [ z ] but not in R [ z ]. Definition 4.3.9. The field F is called algebraically closed if every p F [ z ] splits. 51 4.4 The Fundamental Theorem of Algebra Definition 4.4.1. A function f : C C is entire if there is a power series representation f ( z ) = X n =0 a n z n where a n C for all n Z 0 that converges for all z C . Examples 4.4.2. (1) Every polynomial in C [ z ] is entire as its power series representation is itself. (2) f ( z ) = e z is entire as e z = X j =0 z n n ! (3) f ( z ) = cos( z ) is entire as cos( z ) = X j =0 ( - 1) n z 2 n (2 n )! (4) f ( z ) = sin( z ) is entire as sin( z ) = X j =0 ( - 1) n z 2 n +1 (2 n + 1)! In order to prove the Fundamental Theorem of Algebra, we will need a theorem from complex analysis. We state it without proof. Theorem 4.4.3 (Liouville) . Every bounded, entire function f : C C is constant. Theorem 4.4.4 (Fundamental Theorem of Algebra) . Every nonconstant polynomial in C [ x ] has a root. Proof. Let p be a polynomial in C [ z ]. If p has no roots, then 1 /p is entire and bounded. By Liouville’s Theorem, 1 /p is constant, so p is constant. Remark 4.4.5 . Let p C [ z ] be a nonconstant polynomial, and let n = deg( p ). Then by 4.4.4, p has a root, say μ 1 , so there is a p 1 C [ z ] such that p ( z ) = ( z - μ 1 ) p 1 ( z ). If p 1 is nonconstant, then apply 4.4.4 again to see p 1 has a root μ 2 , so there is a p 2 C [ z ] with deg( p 2 ) = n - 2 such that p ( z ) = ( z - μ 1 )( z - μ 2 ) p 2 ( z ). Repeating this process, we see that p n must be constant as a polynomial has at most n roots. Hence there are μ 1 , . . . , μ n C  #### You've reached the end of your free preview.

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• Fall '08
• GUREVITCH
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