examples of minimal polynomials

examples of minimal polynomials

Note that

2

4

is algebraic over the fields

ℚ

and

ℚ

⁢

(

2

)

. The

minimal polynomials

for

2

4

over these fields are

x

4

-

2

and

x

2

-

2

, respectively. Note that

x

4

-

2

is

irreducible

over

ℚ

by using Eisenstein’s criterion and Gauss’s lemma (

http://planetmath.org/GausssLemmaII

) (see this entry (

http://planetmath.org/AlternativeProofThatSqrt2IsIrrational

) for more details), and

x

2

-

2

is irreducible over

ℚ

⁢

(

2

)

since it is a

quadratic polynomial

and neither of its roots (

2

4

and

-

2

4

) are in

ℚ

⁢

(

2

)

.

A common method for constructing minimal polynomials for numbers that are

expressible

over

ℚ

is “backwards ”: 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 example is

2

4

:

x

=

2

4

x

4

=

2

x

4

-

2

=

0

This method will be further demonstrated with three more examples: One for

1

+

5

2

, one for

1

+

ω

5

where

ω

5

is a fifth

root of unity

, and one for

2

3

+

3

3

.

x

=

1

+

5

2

2

⁢

x

=

1

+

5

2

⁢

x

-

1

=

5

(

2

⁢

x

-

1

)

2

=

5

4

⁢

x

2

-

4

⁢

x

+

1

=

5

4

⁢

x

2

-

4

⁢

x

-

4

=

0

x

2

-

x

-

1

=

0

x

=

1

+

ω

5

x

-

1

=

ω

5

(

x

-

1

)

5

=

1

x

5

-

5

⁢

x

4

+

10

⁢

x

3

-

10

⁢

x

2

+

5

⁢

x

-

1

=

1

x

5

-

5

⁢

x

4

+

10

⁢

x

3

-

10

⁢

x

2

+

5

⁢

x

-

2

=

0

x

=

2

3

+

3

3

x

3

=

2

+

3

⁢

2

2

⋅

3

3

+

3

⁢

2

⋅

3

2

3

+

3

x

3

-

5

=

3

⁢

6

3

⁢

(

2

3

+

3

3

)

x

3

-

5

=

3

⁢

6

3

⁢

x

(

x

3

-

5

)

3

=

27

⋅

6

⁢

x

3

x

9

-

3

⋅

5

⁢

x

6

+

3

⋅

25

⁢

x

3

-

125

=

162

⁢

x

3

x

9

-

15

⁢

x

6

-

87

⁢

x

3

-

125

=

0

Since

x

2

-

x

-

1

is a quadratic and has no roots in

ℚ

, it is irreducible over

ℚ

. Thus, it is the minimal polynomial over

ℚ

for

1

+

5

2

.

On the other hand,

x

5

-

5

⁢

x

4

+

10

⁢

x

3

-

10

⁢

x

2

+

5

⁢

x

-

2

factors over

ℚ

as

(

x

-

2

)

⁢

(

x

4

-

3

⁢

x

3

+

4

⁢

x

2

-

2

⁢

x

+

1

)

. Since

1

+

ω

5

is not a root of

x

-

2

, it must be a root of

x

4

-

3

⁢

x

3

+

4

⁢

x

2

-

2

⁢

x

+

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

+

ω

5

over

ℚ

. Since

ℚ

⁢

(

1

+

ω

5

)

=

ℚ

⁢

(

ω

5

)

,

deg

m

(

x

)

=

[

ℚ

(

1

+

ω

5

)

:

ℚ

]

=

[

ℚ

(

ω

5

)

:

ℚ

]

=

φ

(

5

)

=

4

=

deg

(

x

4

-

3

x

3

+

4

x

2

-

2

x

+

1

)

. (Here

φ

denotes the

Euler totient function

.) Since

m

⁢

(

x

)

divides

x

4

-

3

⁢

x

3

+

4

⁢

x

2

-

2

⁢

x

+

1

and they have the same degree, it follows that

m

⁢

(

x

)

=

x

4

-

3

⁢

x

3

+

4

⁢

x

2

-

2

⁢

x

+

1

.

It turns out that

x

9

-

15

⁢

x

6

-

87

⁢

x

3

-

125

is irreducible over

ℚ

. (This can be proven in a manner as above. Note that

[

ℚ

(

2

3

+

3

3

)

:

ℚ

]

=

9

.) Thus, it is the minimal polynomial over

ℚ

for

2

3

+

3

3

.

Title

examples of minimal polynomials

Canonical name

ExamplesOfMinimalPolynomials

Date of creation

2013-03-22 16:55:18

Last modified on

2013-03-22 16:55:18

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

13

Author

Wkbj79 (1863)

Entry type

Example

Classification

msc 12F05

Classification

msc 12E05

Related topic

IrreduciblePolynomialsObtainedFromBiquadraticFields

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