21. Referring to Example 50.9, show that G(Q(V2, iv3)/Q(iv3)) = < Z,, +>.
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Section 50, Number 21
Transcribed Image Text:21. Referring to Example 50.9, show that G(Q(V2, iV3)/Q(iV3)) = < Z,, +>.
![50.9 Example Let 2 be the real cube root of 2, as usual. Now x - 2 does not split in Q(/2), for
QV2) < R and only one zero of x³ – 2 is real. Thus x³ – 2 factors in (Q(/2))[x] into
V2 and an irreducible quadratic factor. The splitting field E of x³ – 2
a linear factor x
over Q is therefore of degree 2 over Q(V2). Then
[E: Q] = [E : Q(2)][Q(/2) : Q] = (2)(3) = 6.
We have shown that the splitting field over Q of x-2 is of degree 6 over Q.
We can verify by cubing that
and
are the other zeros of x'-2 in C. Thus the splitting field E of x-2 over Q is
Q/2, i/3). (This is not the same field as Q(2, i, V3), which is of degree 12 over Q.)
Further study of this interesting example is left to the exercises (see Exercises 7, 8, 9,
16, 21, and 23).
|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e7c908-505f-4c35-8ac4-e541f2ab69b2%2F156fa328-727c-44e4-9280-5c34f1524d0e%2F73c759s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:50.9 Example Let 2 be the real cube root of 2, as usual. Now x - 2 does not split in Q(/2), for
QV2) < R and only one zero of x³ – 2 is real. Thus x³ – 2 factors in (Q(/2))[x] into
V2 and an irreducible quadratic factor. The splitting field E of x³ – 2
a linear factor x
over Q is therefore of degree 2 over Q(V2). Then
[E: Q] = [E : Q(2)][Q(/2) : Q] = (2)(3) = 6.
We have shown that the splitting field over Q of x-2 is of degree 6 over Q.
We can verify by cubing that
and
are the other zeros of x'-2 in C. Thus the splitting field E of x-2 over Q is
Q/2, i/3). (This is not the same field as Q(2, i, V3), which is of degree 12 over Q.)
Further study of this interesting example is left to the exercises (see Exercises 7, 8, 9,
16, 21, and 23).
|
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