## Chapter 6: Fields: Solved problems

1. Let u be a root of the polynomial x3+3x+3. In Q(u), express (7-2u+u2)-1 in the form a+bu+cu2.

Solution: Dividing x3+3x+3 by x2-2x+7 gives the quotient x+2 and remainder -11. Thus u3+3u+3 = (u+2)(u2-2u+7) - 11, and so (7-2u+u2)-1 = (2+u)/11 = (2/11) + (1/11)u.

2. (a) Show that Q( + i ) = Q( , i ).

Solution: Let u = + i. Since (+i)(-i)=2-i2=3, we have -i = 3 (+i)-1 in Q(u), and it follows easily that is in Q(u) and i is in Q(u), so Q(, i) Q(u). The reverse inclusion is obvious.

(b) Find the minimal polynomial of + i over Q.

Solution: We have Q Q () Q (,i). Thus [Q () : Q] = 2 since is a root of a polynomial of degree 2 but is not in Q. We have [Q (,i) : Q ()] = 2 since i is a root of a polynomial of degree 2 over Q () but is not in Q (). Thus [Q (+i) : Q] = 4, and so the minimal polynomial for + i must have degree 4.

Since u = + i, we have u - i = , u2 - 2iu + i2 = 2, and u2 - 3 = 2iu. Squaring again and combining terms gives u4-2u2+9 = 0. Thus the minimal polynomial for +i is x4 - 2x2 + 9.

3. Find the minimal polynomial of 1 + over Q.

Solution: Let x = 1 + . Then x-1 = , and so (x-1)3 = 2, which yields x3-3x2+3x-1 = 2, and therefore x3 -3x2 +3x -3 = 0. Eisenstein's criterion (with p=3) shows that x3 -3x2 +3x -3 is irreducible over Q, so this is the required minimal polynomial.

4. Show that x3 + 6x2 -12x + 2 is irreducible over Q, and remains irreducible over Q().

Solution: Eisenstein's criterion works with p=2. Since x5 - 2 is also irreducible by Eisenstein's criterion, [Q (): Q] = 5. If x3 + 6x2 -12x + 2 could be factored over Q(), then it would have a linear factor, and so it would have a root in Q(). This root would have degree 3 over Q, and that is impossible since 3 is not a divisor of 5.

5. Find a basis for Q(, ) over Q.

Solution: The set { 1, , } is a basis for Q ( ) over Q, and since this extension has degree 3, the minimal polynomial x2-5 of remains irreducible in the extension Q ( ). Therefore { 1, } is a basis for Q (, ) over Q (), and so the proof of Theorem 6.2.4 shows that the required basis is { 1, , , , , }.

6. Show that [Q ( + ): Q] = 6.

Solution: The set { 1, , } is a basis for Q ( ) over Q, and since this extension has degree 3, the minimal polynomial x2-2 of remains irreducible over the extension Q ( ). Thus { 1, , , , , } is a basis for Q (,) over Q, and this extension contains u = + . It follows that u has degree 2, 3, or 6 over Q.

We will show that u cannot have degree 3. If + is a root of a polynomial ax3+bx2+cx+d in Q [x], then

a ( + )3 + b ( + )2 + c ( + ) + d =
a ( 2 + 6 + 3 + 5 ) + b ( 2 + 2 + ) + c ( + ) + d =
( 5a + 2b + d ) · 1 + ( 6a + c ) + b + ( 2a + c ) + 2b + 3a = 0.

Since { 1, , , , , } are linearly independent over Q, it follows immediately that a=b=0, and then c=d=0 as well, so + cannot satisfy a nonzero polynomial of degree 1, 2, or 3 over Q. We conclude that [Q( + ): Q] = 6.

7. Find [Q( + 3): Q].

Solution: Let u = + 3. Since u = ( + 3)()3, it follows that u is in Q ( ). Since x7 - 2 is irreducible over Q by Eisenstein's criterion, we have [Q ( ) : Q] = 7, and then u must have degree 7 over Q since [Q (u) : Q] is a divisor of [Q () : Q].

8. Find the degree of + i over Q. Does belong to Q ( + i )?

Solution: Let = + i, so that - i = . Then ( - i)3 = 2, so we have 3 - 3i2 +3i2 -i3 = 2, or 3 - 3i2 -3 +i = 2. Solving for i we get i = (3 - 3 - 2)/(32 - 1), and this shows that i is in Q ( + i ). It follows immediately that is in Q ( + i ), and so Q ( + i ) = Q ( , i ).

Since x3-2 is irreducible over Q, the number has degree 3 over Q. Since x2+1 is irreducible over Q, we see that i has degree 2 over Q. Therefore [Q ( + i ) : Q] 6. On the other hand,
[Q ( + i ) : Q] = [Q ( + i ) : Q ()] [Q () : Q]
and
[Q ( + i ) : Q] = [Q ( + i ) : Q (i)] [Q ( i) : Q]
so [Q ( + i ) : Q] must be divisible by 2 and 3. Therefore [Q ( + i ) : Q] = 6.

Finally, has degree 4 over Q since x4-2 is irreducible over Q, so it cannot belong to an extension of degree 6 since 4 is not a divisor of 6.