*Solution:*
Dividing x^{3}+3x+3 by x^{2}-2x+7 gives the quotient x+2 and remainder -11.
Thus u^{3}+3u+3 = (u+2)(u^{2}-2u+7) - 11,
and so (7-2u+u^{2})^{-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-i^{2}=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 = , u^{2} - 2iu + i^{2} = 2, and u^{2} - 3 = 2iu.
Squaring again and combining terms gives u^{4}-2u^{2}+9 = 0.
Thus the minimal polynomial for +i is x^{4} - 2x^{2} + 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 x^{3}-3x^{2}+3x-1 = 2,
and therefore x^{3} -3x^{2} +3x -3 = 0.
Eisenstein's criterion (with p=3) shows that
x^{3} -3x^{2} +3x -3 is
irreducible
over **Q**, so this is the required minimal polynomial.

**4.**
Show that x^{3} + 6x^{2} -12x + 2 is
irreducible
over **Q**, and remains irreducible over **Q**().

*Solution:*
Eisenstein's criterion works with p=2.
Since x^{5} - 2 is also irreducible by
Eisenstein's criterion,
[**Q** (): **Q**] = 5.
If x^{3} + 6x^{2} -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 x^{2}-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 x^{2}-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 ax^{3}+bx^{2}+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 x^{7} - 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} - 3i^{2} +3i^{2} -i^{3} = 2, or
^{3} - 3i^{2} -3 +i = 2.
Solving for i we get i = (^{3} - 3 - 2)/(3^{2} - 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 x^{3}-2 is
irreducible
over **Q**,
the number has degree 3 over **Q**.
Since x^{2}+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 x^{4}-2 is
irreducible
over **Q**, so it cannot belong to an extension of degree 6
since 4 is not a divisor of 6.