College Algebra Exam Review 319

# College Algebra Exam Review 319 - roots of f.x If ı 2 K...

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7.4. SPLITTING FIELD OF A CUBIC POLYNOMIAL 329 K.˛ 1 / ± E , it follows from the multiplicativity of dimension (Proposition 7.3.1 ) that 3 divides the dimension of E over K . Recall the element ı D 1 ² ˛ 2 /.˛ 1 ² ˛ 3 /.˛ 2 ² ˛ 3 / 2 E ; since ı 2 D ² 4p 3 ² 27q 2 2 K , either ı 2 K or K.ı/ is an extension ﬁeld of dimension 2 over K . In the latter case, since K ± K.ı/ ± E , it follows that 2 also divides dim K .E/ . We will show that there are only two possibilities: 1. ı 2 K and dim K .E/ D 3 , or 2. ı 62 K and dim K .E/ D 6 . We’re going to do some algebraic tricks to solve for ˛ 2 in terms of ˛ 1 and ı . The identity P i ˛ i D 0 gives: ˛ 3 D ² ˛ 1 ² ˛ 2 : (7.4.1) Eliminating ˛ 3 in P i<j ˛ i ˛ j D p gives ˛ 2 2 D ² ˛ 2 1 ² ˛ 1 ˛ 2 ² p: (7.4.2) Since f.˛ i / D 0 , we have ˛ 3 i D ² i ² q .i D 1;2/: (7.4.3) In the Exercises, you are asked to show that ˛ 2 D ı C 1 p C 3q 2 ± 1 2 C p ² : (7.4.4) Proposition 7.4.1. Let K be a subﬁeld of C , let f.x/ D x 3 C px C q 2 KŒxŁ an irreducible cubic polynomial, and let E denote the splitting ﬁeld of f.x/ in C . Let ı D 1 ² ˛ 2 /.˛ 1 ² ˛ 3 /.˛ 2 ² ˛ 3 / , where ˛ i are the
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Unformatted text preview: roots of f.x/ . If ı 2 K , then dim K .E/ D 3 . Otherwise, dim K .E/ D 6 . Proof. Suppose that ı 2 K . Then Equation ( 7.4.4 ) shows that ˛ 2 and, therefore, also ˛ 3 is contained in K.˛ 1 / . Thus E D K.˛ 1 ;˛ 2 ;˛ 3 / D K.˛ 1 / , and E has dimension 3 over K . On the other hand, if ı 62 K , then we have seen that dim K .E/ is divisible by both 2 and 3, so dim K .E/ ³ 6 . Consider the ﬁeld extension K.ı/ ± E . If f.x/ were not irreducible in K.ı/ŒxŁ , then we would have dim K.ı/ .E/ ´ 2 , so dim K .E/ D dim K .K.ı// dim K.ı/ .E/ ´ 4; a contradiction. So f.x/ must remain irreducible in K.ı/ŒxŁ . But then it follows from the previous paragraph (replacing K by K.ı/ ) that dim K.ı/ .E/ D 3 . Therefore, dim K .E/ D 6 . n...
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