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examples of minimal polynomials∗
Wkbj79†
2013-03-21 22:31:48
√
√
4
Note that
2 is algebraic over the fields Q and √
Q( 2). The minimal poly√
4
nomials for 2 over these fields are x4 − 2 and x2 − 2, respectively. Note that
x4 − 2 is irreducible over Q by using Eisenstein’s
criterion and Gauss’s
√ lemma
√
over
Q( 2) since
(see this entry for more details), and x2 − 2 is irreducible
√
√
√ it
is a quadratic polynomial and neither of its roots ( 4 2 and − 4 2) are in Q( 2).
A common method for constructing minimal polynomials for numbers that
are expressible over Q is “backwards algebra”: The number can be set equal to
x, and the equation can be algebraically manipulated until a monic polynomial
in Q[x] is equal to 0. Finally, if the monic polynomial is not irreducible, then it
can be factored into irreducible polynomials Q[x], and
√ the original number will
be a root of one of these. A very simple example is 4 2:
√
x = 42
x4 = 2
4
x −2 =0
This√method will be further demonstrated with three more examples: One
√
√
1+ 5
for
, one for 1 + ω5 where ω5 is a fifth root of unity, and one for 3 2 + 3 3.
2
√
1+ 5
x =
2√
2x = √
1+ 5
2x − 1 = 5
(2x − 1)2 = 5
4x2 − 4x + 1 = 5
4x2 − 4x − 4 = 0
x2 − x − 1 = 0
∗ hExamplesOfMinimalPolynomialsi created: h2013-03-21i by: hWkbj79i version: h39184i
Privacy setting: h1i hExamplei h12F05i h12E05i
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compatible with the CC-BY-SA license.
1
x = 1 + ω5
x − 1 = ω5
(x − 1)5 = 1
5
4
3
2
x − 5x + 10x − 10x + 5x − 1 = 1
x5 − 5x4 + 10x3 − 10x2 + 5x − 2 = 0
√
√
x = 3 2 +√3 3
√
3
3
x3 = 2 √
+ 3 √22 · 3√+ 3 2 · 32 + 3
3
3
3
x3 − 5 = 3 √6( 2 + 3)
x3 − 5 = 3 3 6 x
3
(x − 5)3 = 27 · 6x3
9
6
x − 3 · 5x + 3 · 25x3 − 125 = 162x3
x9 − 15x6 − 87x3 − 125 = 0
Since x2 − x − 1 is a quadratic and has no roots in√Q, it is irreducible over
1+ 5
.
Q. Thus, it is the minimal polynomial over Q for
2
5
4
3
2
On the other hand, x − 5x + 10x − 10x + 5x − 2 factors over Q as
(x − 2)(x4 − 3x3 + 4x2 − 2x + 1). Since 1 + ω5 is not a root of x − 2, it must be a
root of x4 − 3x3 + 4x2 − 2x + 1. Moreover, this polynomial must be irreducible.
This fact can be proven in the following manner: Let m(x) be the minimal
polynomial for 1 + ω5 over Q. Since Q(1 + ω5 ) = Q(ω5 ), deg m(x) = [Q(1 + ω5 ) :
Q] = [Q(ω5 ) : Q] = ϕ(5) = 4 = deg(x4 − 3x3 + 4x2 − 2x + 1). (Here ϕ denotes
the Euler totient function.) Since m(x) divides x4 − 3x3 + 4x2 − 2x + 1 and they
have the same degree, it follows that m(x) = x4 − 3x3 + 4x2 − 2x + 1.
It turns out that x9 − 15x6 − 87x3 − 125 is irreducible
over
Q. (This can be
√
√
3
3
2
+
3)
:
Q] = 9.) Thus,
proven in a similar manner as above. Note
that
[Q(
√
√
it is the minimal polynomial over Q for 3 2 + 3 3.
2