### Minimal Polynomials of Algebraic Elements

```OUTLINE
-
Definitions
Rings
Fields
Field Extensions
Algebraic
Minimal Polynomials
-
First Isomorphism Theorem
-
Main Theorem(for simple field extensions)
-
Example of Main Theorem
-
What if my field extension is not simple?!
WHAT IS A RING ?
A ring is a set R equipped with two binary operations called addition and multiplication
such that:
- (R,+) is an abelian (commutative) group
- (R, x) has the following properties:
- Closed
- Associative
- There is a multiplicative identity “1”
- Multiplication distributes over the addition
Examples:
-Z
, Z4 , R , Q
WHAT IS A FIELD ?
A field is commutative ring whose nonzero elements form a group under
multiplication
So a field is commutative ring whose nonzero elements each have a multiplicative
inverse
Examples:
- Zp where p is prime, R, Q, C
Non-examples:
- Zn where n is not prime
WHAT IS A FIELD
EXTENSION ?
Let K be a field. If a set k⊆ K is closed under the field operations and inverses in K (i.e.
k is a subfield of K) then we call K an extension field of k. We say K/k, “K over k,” is
a field extension.
Example:
- R and Q are both fields where Q ⊆ R. Therefore, R is an extension field of Q. We
could state this as: R/Q is a field extension. The same relationship holds for R ⊆C.

Extension field of R
Extension field of Q and
Subfield of C
Subfield of R
CONSTRUCTING AN EXTENSION FIELD
Consider a base field k. Adjoin with it an element α ∈ k to construct k(α), the smallest
field that contains the base field k and also α.
Example:
- If we start with the field Q, we can build the field extension Q(√2)/Q by adjoining
√2 to Q where Q(√2) = {a + b√2 | a, b ∈ Q} is an extension field.
Note: we call any extension of the form k(α) a simple extension field because only one
(√2)

Extension
Field (K)
Subfield (k)
WHAT IS AN
ALGEBRAIC ELEMENT ?
If K is a field extension of k, then an element α∈ K is called algebraic over k if there
exists some non-zero polynomial g(x) ∈ k[x] such that g(α) = 0.
Example:
- Consider Q(√2), an extension field of Q. √2 is algebraic over Q since it is a root of
−  ∈ Q[x].
Non-Example:
- Consider Q(π). π is a root of x – π, but this polynomial is not in Q[x]. So, π is not
algebraic over Q.
WHAT IS A MINIMAL
POLYNOMIAL ?
Let α ∈ K be algebraic over k. The minimal polynomial of α is the monic polynomial p
∈k[x] of least degree such that p(α) = 0.
Proposition: The minimal polynomial is irreducible over k and any other non-zero
polynomial f(x) such that f(α) = 0 must be a multiple of p.
Example:
α = √2 element of Q(√2) where √2 is algebraic over Q. The minimal polynomial of
√2 is  − .
WHAT ARE QUOTIENT
RINGS ?
A quotient ring is constructed out of a given ring R by a process of “moding out” by an
ideal I ⊆ R, denoted R/I.
Ex.1
/  ≅
In general, /  ≅
Ex.2
/  +  ≅
where  +  is the ideal
Note, when you have a principal ideal, it is generator is the minimal polynomial
CONSEQUENCE OF THE
FIRST ISOMORPHISM THEOREM
Given K/k, with  ∈  algebraic over k and the minimal polynomial  of , consider
the homomorphism
:

x

N: ker  =
By the First Isomorphism Theorem,
/  ≅ ()
Domain
Kernel
Range
Given K = k(α), with α algebraic over k, how do we
compute the minimal polynomial of any β ∈ K ?
Main Theorem: Let K/k be a field extension, and let α ∈ K be algebraic over k.
Let  ∈ [] be the minimal polynomial of α over k. Let 0 ≠  ∈ k(α), say
+ 1  + ⋯ +
=
0 + 1  + ⋯ +
where  ,  ∈ , 0 ≤  ≤ , 0 ≤  ≤ . Let   =  + 1  + ⋯ +    and
() = 0 + 1  + ⋯ +    be the corresponding polynomials in
[]. Consider the ideal  =< ,  −  > of [, ]. Then the minimal
polynomial of  over k is the monic polynomial that generates the ideal    .
Note: since   /  ≅ (), we can compute in   /
Proof.
Since []/  is a field and   ≠ 0, there is a polynomial  ∈   such that
≡ 1 (  ) that is  − 1 ∈
Let ℎ = . Note that ℎ  = .
Now consider the map  ∶
/

ℎ+
≅
()

is in the kernel of  if and only if   = 0. Therefore, the minimal polynomial of
will be in the kernel of .
We now find the generator of the kernel of  because
this will be minimal polynomial !!!
…
By another theorem, the kernel of  is ,  − ℎ
are considering the ideal  =< ,  −  >.
[] . Recall that we
So, if we can show ,  − ℎ =< ,  −  > then we know that the kernel of the map  is
,  −  [] , namely the minimal of  is the monic polynomial that generates the ideal
, and we are done!
To show this, use containment: ,  −  ⊇ ,  − ℎ
− ℎ =  −  ≡   −

− ℎ ∈ ,  −
Now show: ,  −  ⊆ ,  − ℎ
−  ≡   −  =   − ℎ
−  ∈ ,  − ℎ
,  − ℎ = ,  −
RESULTS!
The minimal polynomial of  over k is the monic polynomial that
generates the ideal  []
This gives us an algorithm for finding the minimal polynomial:
Given  and  as in the theorem, compute the Groebner basis G
for the ideal ,  −  of k[x,y] with respect to the lex term
ordering x > y. The polynomial in G which is in y alone is the
minimal polynomial of .
LET’S FIND THE MINIMAL POLYNOMIAL!
Consider the field Q() , where  is a root of the irreducible polynomial:  5 −  − 2
We want to find the minimal polynomial for an element  =
()
()
=
1−  −2 3

∈ ()
Let’s construct the ideal as the theorem/algorithm
= , () − (x) =  5 −  − 2,  − (1 −  − 2 3 ) =  5 −  − 2,  + 2 3 +
minimal polynomial of !
THIS IS TRUE !
Recall, β =
1− α −2α3
α
and α is a root of x 5 − x − 2.
5 +
β5 +
1− α −2α3 5
)
α
(
+
11 4
+ 4 3 − 5 2 + 95y + 259
2
11 4
β
2
11 1− α −2α3 4
(
)
2
α
+ 4β3 − 5β2 + 95β + 259 =
+ 4(
1− α −2α3 3
)
α
− 5(
1− α −2α3 2
)
α
1− α −2α3
+ 95(
α
) + 259 =
…
=0
So, β is a root of the (irreducible over k) polynomial

+
+   −   +  +

This is the minimal polynomial for  over k.
WHAT IF THE FIELD EXTENSION IS NONSIMPLE?
As defined, a simple field extension can be written with only one element adjoined to
the base field in order to make up the extension field.
A non-simple field extension has multiple elements adjoined to the base field to make
up the extension field.
Examples:
- Q( , )
- R/Q
In a non-simple extension field, we want to extend our above theorem in order to find
the minimal polynomial of any element  of   ,  , … ,
Given K =   ,  , … ,  , with  ’s algebraic over k, how do
we compute the minimal polynomial of any β ∈ K ?
Main Theorem Revised: Let K=  1 , … ,  be a field extension, and let  ’s
∈ K be algebraic over k. Let  ∈  1 , … , −1  be the minimal polynomial
of  over  1 , … , −1 . Let 0 ≠  ∈  1 , … , −1 , say
(1 , … ,  )
=
(1 , … ,  )
where ,  ∈  1 , … . ,  . Consider the ideal  =< 1 , … ,  ,  −  >
contained in [1 , … ,  , ]. Then the minimal polynomial of  over k is the
monic polynomial that generates the ideal    .
Note: For  = 2, … ,  and  ∈  1 , … , −1  , we let  be any polynomial in
1 , … ,  such that  1 , … , −1 ,  = .
```