Minimal polynomials are key to understanding algebraic elements in field extensions. They help determine if an element is algebraic and provide a way to construct field extensions. The degree of a equals the dimension of the it creates.
, determined by the minimal polynomial, tells us how "complex" an element is over a field. It's crucial for understanding the structure of field extensions and plays a vital role in solving polynomial equations and studying field automorphisms.
Minimal Polynomials for Algebraic Elements
Definition and Properties
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An element α in an extension field E of a field F is algebraic over F if there exists a non-zero polynomial f(x) in F[x] such that f(α)=0
The minimal polynomial of an algebraic element α over a field F is the unique monic polynomial m(x) in F[x] of least degree such that m(α)=0
The minimal polynomial of an algebraic element is irreducible over the base field
If α is algebraic over F, then F(α) is the smallest subfield of E containing both F and α (example: Q(2) is the smallest subfield of R containing both Q and 2)
If α is transcendental over F, then there is no non-zero polynomial f(x) in F[x] such that f(α)=0 (example: π is transcendental over Q)
Algebraic Elements and Field Extensions
If α is algebraic over F with minimal polynomial m(x), then F(α) is isomorphic to the quotient ring F[x]/(m(x))
The degree of the field extension F(α) over F is equal to the algebraic degree of α over F
If α and β are algebraic over F with minimal polynomials m(x) and n(x) respectively, then α+β and αβ are also algebraic over F
The minimal polynomial of α+β divides the polynomial m(x+y) in F[x,y]
The minimal polynomial of αβ divides the polynomial m(xy) in F[x,y]
If α is algebraic over F and β is algebraic over F(α), then β is algebraic over F
The minimal polynomial of β over F divides the polynomial obtained by substituting the minimal polynomial of α into the minimal polynomial of β over F(α)
Computing Minimal Polynomials
Finding the Minimal Polynomial
To find the minimal polynomial of an algebraic element α over a field F, first find a non-zero polynomial f(x) in F[x] such that f(α)=0
Factor f(x) into irreducible factors over F. The minimal polynomial m(x) will be one of these irreducible factors
Substitute α into each irreducible factor. The factor that evaluates to 0 is the minimal polynomial m(x)
If necessary, divide m(x) by its leading coefficient to obtain a monic polynomial
Verify that m(α)=0 and that no polynomial of lower degree in F[x] has α as a root
Examples
Find the minimal polynomial of 2 over Q:
The polynomial x2−2 has 2 as a root, so f(x)=x2−2
f(x) is already irreducible over Q, so m(x)=x2−2
m(x) is monic, and no polynomial of lower degree in Q[x] has 2 as a root
Find the minimal polynomial of i+2 over Q:
The polynomial x4−2x2+9 has i+2 as a root, so f(x)=x4−2x2+9
f(x) factors into (x2−2x−3)(x2+2x−3) over Q
Substituting i+2 into each factor, we find that m(x)=x2−2x−3
m(x) is monic, and no polynomial of lower degree in Q[x] has i+2 as a root
Algebraic Degree and Minimal Polynomials
Definition and Properties
The algebraic degree of an element α over a field F is the degree of its minimal polynomial m(x) over F
If the minimal polynomial of α over F is linear, then α is in F, and the algebraic degree is 1
If the minimal polynomial of α over F has degree n>1, then the algebraic degree of α over F is n
The algebraic degree of α over F is equal to the dimension of F(α) as a vector space over F
Examples
The minimal polynomial of 2 over Q is x2−2, so the algebraic degree of 2 over Q is 2
The dimension of Q(2) as a vector space over Q is also 2, with basis {1,2}
The minimal polynomial of i+2 over Q is x2−2x−3, so the algebraic degree of i+2 over Q is 2
The dimension of Q(i+2) as a vector space over Q is also 2, with basis {1,i+2}
Minimal Polynomials vs Field Extensions
Isomorphism and Degree
If α is algebraic over F with minimal polynomial m(x), then F(α) is isomorphic to the quotient ring F[x]/(m(x))
The degree of the field extension F(α) over F is equal to the algebraic degree of α over F
This follows from the isomorphism between F(α) and F[x]/(m(x)), as the dimension of F[x]/(m(x)) as a vector space over F is equal to the degree of m(x)
Algebraic Elements and Field Extensions
If α and β are algebraic over F with minimal polynomials m(x) and n(x) respectively, then α+β and αβ are also algebraic over F
The minimal polynomial of α+β divides the polynomial m(x+y) in F[x,y]
The minimal polynomial of αβ divides the polynomial m(xy) in F[x,y]
If α is algebraic over F and β is algebraic over F(α), then β is algebraic over F
The minimal polynomial of β over F divides the polynomial obtained by substituting the minimal polynomial of α into the minimal polynomial of β over F(α)
This result allows us to build larger field extensions by adjoining algebraic elements step by step (example: Q(2,3) can be constructed by first adjoining 2 to Q, then adjoining 3 to Q(2))