Examples
The following numbers are algebraic:
3,
3 + 1, 3 + 2, 21/3 + 21/2.
For, they are zeros of the following polynomials.
We show how to find such a polynomial for
= 3 + 2
= 21/3 + 21/2
The key is to find a Q-linear relation between the powers of
. First form 2 = 5 + 26. The
three elements 1, , and 2 are written as Q-linear
combinations of the independent elements 1, 2, 3, 6. Because we cannot yet expect a linear relation,
we calculate the next power: 3 =
93 + 112. Still no
linear relation, so we continue 4
= 49 + 206. But now 4 = 102 - 1, so
satisfies X4 - 10X2 + 1 =
0.
. First form 2 = 5 + 26. The
three elements 1, , and 2 are written as Q-linear
combinations of the independent elements 1, 2, 3, 6. Because we cannot yet expect a linear relation,
we calculate the next power: 3 =
93 + 112. Still no
linear relation, so we continue 4
= 49 + 206. But now 4 = 102 - 1, so
satisfies
When computing powers of
, we find Q-linear combinations of powers of
21/6. Therefore, we determine a 7×6 matrix whose
rows are the powers of ,
written out with respect to the basis 1, 21/6, ..., 25/6:
f =
X6 - 6X4 - 4X3 + 12X2 - 24X - 4
, we find Q-linear combinations of powers of
21/6. Therefore, we determine a 7×6 matrix whose
rows are the powers of ,
written out with respect to the basis 1, 21/6, ..., 25/6:
|
|
|
Next, we look for a linear relation between the rows. This amounts to finding a vector in the kernel of the transposed matrix. As a row vector, this is (-4,-24,12,-4,-6,0,1), which means that the polynomial
is as required.
It is straightforward now to verify
f() = 0.