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.
𝑪
𝑹
𝑸
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
element is adjoined.
𝑸(√2)
𝑸
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
𝒙𝟐 − 𝟐 ∈ 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 𝒙𝟐 − 𝟐.
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
𝒁/ 𝟑 ≅ 𝒁𝟑
In general, 𝒁/ 𝒑 ≅ 𝒁𝒑
Ex.2
𝑹 𝒙 / 𝒙𝟐 + 𝟏 ≅ 𝑪
where 𝒙𝟐 + 𝟏 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 𝛼 ∈ 𝐾 algebraic over k and the minimal polynomial 𝑝 of 𝜶, consider
the homomorphism
𝜑: 𝑘 𝑥
𝑘 𝛼
x
𝛼
N𝐨𝐭𝐞: ker 𝜑 = 𝑝
By the First Isomorphism Theorem,
𝑘 𝑥 / 𝑝 ≅ 𝑘(𝛼)
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 𝑝 ∈ 𝑘[𝑥] be the minimal polynomial of α over k. Let 0 ≠ 𝛽 ∈ k(α), say
𝑎𝑜 + 𝑎1 𝛼 + ⋯ + 𝑎𝑛 𝛼 𝑛
𝛽=
𝑏0 + 𝑏1 𝛼 + ⋯ + 𝑏𝑚 𝛼 𝑚
where 𝑎𝑖 , 𝑏𝑗 ∈ 𝑘, 0 ≤ 𝑖 ≤ 𝑛, 0 ≤ 𝑗 ≤ 𝑚. Let 𝑓 𝑥 = 𝑎𝑜 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 and
𝑔(𝑥) = 𝑏0 + 𝑏1 𝑥 + ⋯ + 𝑏𝑚 𝑥 𝑚 be the corresponding polynomials in
𝑘[𝑥]. Consider the ideal 𝐽 =< 𝑝, 𝑔𝑦 − 𝑓 > of 𝑘[𝑥, 𝑦]. Then the minimal
polynomial of 𝛽 over k is the monic polynomial that generates the ideal 𝐽 𝑘 𝑦 .
Note: since 𝑘 𝑥 / 𝑝 ≅ 𝑘(𝛼), we can compute in 𝑘 𝑥 / 𝑝
Proof.
Since 𝑘[𝑥]/ 𝑝 is a field and 𝑔 𝛼 ≠ 0, there is a polynomial 𝑙 ∈ 𝑘 𝑥 such that 𝑔𝑙
≡ 1 (𝑚𝑜𝑑 𝑝 ) that is 𝑔𝑙 − 1 ∈ 𝑝
Let ℎ = 𝑓𝑙. Note that ℎ 𝛼 = 𝛽.
Now consider the map 𝜑 ∶ 𝑘 𝑦
𝑘𝑥/𝑝
𝑦
ℎ+ 𝑝
≅
𝑘(𝛼)
𝛽
𝑞 is in the kernel of 𝜑 if and only if 𝑞 𝛽 = 0. Therefore, the minimal polynomial of 𝛽
will be in the kernel of 𝜑.
We now find the generator of the kernel of 𝝋 because
this will be minimal polynomial !!!
…
By another theorem, the kernel of 𝜑 is 𝑝, 𝑦 − ℎ
are considering the ideal 𝐽 =< 𝑝, 𝑔𝑦 − 𝑓 >.
𝑘[𝑦] . Recall that we
So, if we can show 𝑝, 𝑦 − ℎ =< 𝑝, 𝑔𝑦 − 𝑓 > then we know that the kernel of the map 𝜑 is
𝑝, 𝑔𝑦 − 𝑓 𝑘[𝑦] , namely the minimal of 𝛽 is the monic polynomial that generates the ideal
𝐽 𝑘 𝑦 , and we are done!
To show this, use containment: 𝑝, 𝑔𝑦 − 𝑓 ⊇ 𝑝, 𝑦 − ℎ
𝑦 − ℎ = 𝑦 − 𝑓𝑙 ≡ 𝑙 𝑔𝑦 − 𝑓
𝑚𝑜𝑑 𝑝
𝑦 − ℎ ∈ 𝑝, 𝑔𝑦 − 𝑓
Now show: 𝑝, 𝑔𝑦 − 𝑓 ⊆ 𝑝, 𝑦 − ℎ
𝑔𝑦 − 𝑓 ≡ 𝑔 𝑦 − 𝑓𝑙 = 𝑔 𝑦 − ℎ 𝑚𝑜𝑑 𝑝
𝑔𝑦 − 𝑓 ∈ 𝑝, 𝑦 − ℎ
𝑝, 𝑦 − ℎ = 𝑝, 𝑔𝑦 − 𝑓
RESULTS!
The minimal polynomial of 𝜷 over k is the monic polynomial that
generates the ideal 𝑱 𝒌[𝒚]
This gives us an algorithm for finding the minimal polynomial:
Given 𝜶 and 𝜷 as in the theorem, compute the Groebner basis G
for the ideal 𝑝, 𝑔𝑦 − 𝑓 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 𝜷.
LET’S FIND THE MINIMAL POLYNOMIAL!
Consider the field Q(𝛼) , where 𝛼 is a root of the irreducible polynomial: 𝑥 5 − 𝑥 − 2
We want to find the minimal polynomial for an element 𝛽 =
𝑓(𝛼)
𝑔(𝛼)
=
1− 𝛼 −2𝛼 3
𝛼
∈ 𝑄(𝛼)
Let’s construct the ideal as the theorem/algorithm
𝐽 = 𝑝, 𝑔(𝑥)𝑦 − 𝑓(x) = 𝑥 5 − 𝑥 − 2, 𝑥𝑦 − (1 − 𝑥 − 2𝑥 3 ) = 𝑥 5 − 𝑥 − 2, 𝑥𝑦 + 2𝑥 3 + 𝑥
minimal polynomial of 𝛽!
THIS IS TRUE !
Recall, β =
1− α −2α3
α
and α is a root of x 5 − x − 2.
𝑦5 +
β5 +
1− α −2α3 5
)
α
(
+
11 4
𝑦 + 4𝑦 3 − 5𝑦 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
𝟏𝟏 𝟒
𝐲𝟓 +
𝐲 + 𝟒𝐲 𝟑 − 𝟓𝐲 𝟐 + 𝟗𝟓𝐲 + 𝟐𝟓𝟗
𝟐
This is the minimal polynomial for 𝜷 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( 𝟐, 𝟑)
- 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 𝜷 of 𝒌 𝜶𝟏 , 𝜶𝟐 , … , 𝜶𝒏
Given K = 𝒌 𝜶𝟏 , 𝜶𝟐 , … , 𝜶𝒏 , with 𝛼𝑖 ’s algebraic over k, how do
we compute the minimal polynomial of any β ∈ K ?
Main Theorem Revised: Let K= 𝑘 𝛼1 , … , 𝛼𝑛 be a field extension, and let 𝛼𝑖 ’s
∈ K be algebraic over k. Let 𝑝𝑖 ∈ 𝑘 𝛼1 , … , 𝛼𝑖−1 𝑥𝑖 be the minimal polynomial
of 𝛼𝑖 over 𝑘 𝛼1 , … , 𝛼𝑖−1 . Let 0 ≠ 𝛽 ∈ 𝑘 𝛼1 , … , 𝛼𝑖−1 , say
𝑓(𝛼1 , … , 𝛼𝑛 )
𝛽=
𝑔(𝛼1 , … , 𝛼𝑛 )
where 𝑓, 𝑔 ∈ 𝑘 𝑥1 , … . , 𝑥𝑛 . Consider the ideal 𝐽 =< 𝑝1 , … , 𝑝𝑛 , 𝑔𝑦 − 𝑓 >
contained in 𝑘[𝑥1 , … , 𝑥𝑛 , 𝑦]. Then the minimal polynomial of 𝛽 over k is the
monic polynomial that generates the ideal 𝐽 𝑘 𝑦 .
Note: For 𝑖 = 2, … , 𝑛 and 𝑝 ∈ 𝑘 𝛼1 , … , 𝛼𝑖−1 𝑥𝑖 , we let 𝑝 be any polynomial in
𝑘 𝑥1 , … , 𝑥𝑖 such that 𝑝 𝛼1 , … , 𝛼𝑖−1 , 𝑥𝑖 = 𝑝.