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§ 6.4 Splitting fields: solved problems

1. Find the splitting field over Q for the polynomial x4 + 4.

Solution: We have the factorization x4 + 4 = (x2 + 2x +2)(x2 - 2x +2), where the factors are irreducible by Eisenstein's criterion (using p = 2). The roots are ± 1 ± i, so the splitting field is Q(i), which has degree 2 over Q.

An alternate solution is to solve x4 = -4, To find one root, use DeMoivre's theorem to get 1/ + 1/ i, and then multiply by to get 1+i. The other roots are found by multiplying by the powers of i, because it is a primitive 4th root of unity.

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2. Find the degree of the splitting field over Z2 for the polynomial (x3 + x + 1)(x2 + x + 1).

Solution: The two polynomials are irreducible (you can check that they have no roots). Therefore the splitting field must have subfields of degree 3 and of degree 2, so the degree of the splitting field over Z2 must be 6.

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3. Find the degree [F:Q], where F is the splitting field of the polynomial x3 - 11 over the field Q of rational numbers.

Solution: Letting be the cube root of 11, the roots of the polynomial are , , and 2 , where is a primitive cube root of unity. Since is not real, it cannot belong to Q(). Since is a root of x2 +x +1 and F = Q (, ), we have [F:Q] = 6.

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4. Determine the splitting field over Q for x4 + 2.

Solution: To get the splitting field F, we need to adjoin the 4th roots of -2, which have the form i , where is the 4th root of 2, is a primitive 8th root of unity, and i = 1,3,5,7. To construct these roots we only need to adjoin and i.

To show this, using the polar form cos + i sin of the complex numbers, we can see that
    = / 2 + / 2 i,   3 = - / 2 + / 2 i,
    5 = - / 2 - / 2 i,   and   7 = / 2 - / 2 i.
Thus = + 7 must belong to F, and then the cube of this element, which is 4 , must also belong to F. Therefore belongs to F (which is somewhat surprising) and the square of this element is , so it follows that belongs to F, and therefore i belongs to F. The splitting field is thus Q ( , i), which has degree 8 over Q.

Note: This is the same field as in Example 8.3.2 of the text, which computes the Galois group of x4-2 over Q.

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5. Determine the splitting field over Q for x4 + x2 + 1.

Solution: Be careful here--this polynomial is not irreducible. In fact, x6 - 1 factors in two ways, and provides an important clue. Note that
    x6 - 1 = (x3)2 - 1 = (x3 - 1)(x3 + 1) = (x-1)(x2+x+1)(x+1)(x2-x+1)   and
    x6 - 1 = (x2)3 - 1 = (x2 - 1)(x4 +x2 +1).
Thus x4+x2+1 = (x2+x+1)(x2-x+1), and the roots of the first factor are the primitive 3rd roots of unity, while the roots of the second factor are the primitive 6th roots of unity. Adjoining a root of x2-x+1 gives all 4 roots, and so the splitting field Q() has degree 2 over Q.


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