2.3 Minimal polynomials and algebraic degree

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 minimal polynomial equals the dimension of the field extension it creates.

Algebraic degree, 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

• An element $\alpha$ 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 $f(\alpha) = 0$
• The minimal polynomial of an algebraic element $\alpha$ over a field $F$ is the unique monic polynomial $m(x)$ in $F[x]$ of least degree such that $m(\alpha) = 0$
  • The minimal polynomial of an algebraic element is irreducible over the base field
  • If $\alpha$ is algebraic over $F$, then $F(\alpha)$ is the smallest subfield of $E$ containing both $F$ and $\alpha$ (example: $\mathbb{Q}(\sqrt{2})$ is the smallest subfield of $\mathbb{R}$ containing both $\mathbb{Q}$ and $\sqrt{2}$)
• If $\alpha$ is transcendental over $F$, then there is no non-zero polynomial $f(x)$ in $F[x]$ such that $f(\alpha) = 0$ (example: $\pi$ is transcendental over $\mathbb{Q}$)

Algebraic Elements and Field Extensions

• If $\alpha$ is algebraic over $F$ with minimal polynomial $m(x)$, then $F(\alpha)$ is isomorphic to the quotient ring $F[x]/(m(x))$
• The degree of the field extension $F(\alpha)$ over $F$ is equal to the algebraic degree of $\alpha$ over $F$
• If $\alpha$ and $\beta$ are algebraic over $F$ with minimal polynomials $m(x)$ and $n(x)$ respectively, then $\alpha + \beta$ and $\alpha\beta$ are also algebraic over $F$
  • The minimal polynomial of $\alpha + \beta$ divides the polynomial $m(x + y)$ in $F[x, y]$
  • The minimal polynomial of $\alpha\beta$ divides the polynomial $m(xy)$ in $F[x, y]$
• If $\alpha$ is algebraic over $F$ and $\beta$ is algebraic over $F(\alpha)$, then $\beta$ is algebraic over $F$
  • The minimal polynomial of $\beta$ over $F$ divides the polynomial obtained by substituting the minimal polynomial of $\alpha$ into the minimal polynomial of $\beta$ over $F(\alpha)$

Computing Minimal Polynomials

Finding the Minimal Polynomial

• To find the minimal polynomial of an algebraic element $\alpha$ over a field $F$, first find a non-zero polynomial $f(x)$ in $F[x]$ such that $f(\alpha) = 0$
• Factor $f(x)$ into irreducible factors over $F$. The minimal polynomial $m(x)$ will be one of these irreducible factors
• Substitute $\alpha$ 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 $m(\alpha) = 0$ and that no polynomial of lower degree in $F[x]$ has $\alpha$ as a root

Examples

• Find the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$:
  • The polynomial $x^2 - 2$ has $\sqrt{2}$ as a root, so $f(x) = x^2 - 2$
  • $f(x)$ is already irreducible over $\mathbb{Q}$, so $m(x) = x^2 - 2$
  • $m(x)$ is monic, and no polynomial of lower degree in $\mathbb{Q}[x]$ has $\sqrt{2}$ as a root
• Find the minimal polynomial of $i + \sqrt{2}$ over $\mathbb{Q}$:
  • The polynomial $x^4 - 2x^2 + 9$ has $i + \sqrt{2}$ as a root, so $f(x) = x^4 - 2x^2 + 9$
  • $f(x)$ factors into $(x^2 - 2x - 3)(x^2 + 2x - 3)$ over $\mathbb{Q}$
  • Substituting $i + \sqrt{2}$ into each factor, we find that $m(x) = x^2 - 2x - 3$
  • $m(x)$ is monic, and no polynomial of lower degree in $\mathbb{Q}[x]$ has $i + \sqrt{2}$ as a root

Algebraic Degree and Minimal Polynomials

Definition and Properties

• The algebraic degree of an element $\alpha$ over a field $F$ is the degree of its minimal polynomial $m(x)$ over $F$
  • If the minimal polynomial of $\alpha$ over $F$ is linear, then $\alpha$ is in $F$, and the algebraic degree is 1
  • If the minimal polynomial of $\alpha$ over $F$ has degree $n > 1$, then the algebraic degree of $\alpha$ over $F$ is $n$
• The algebraic degree of $\alpha$ over $F$ is equal to the dimension of $F(\alpha)$ as a vector space over $F$

Examples

• The minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $x^2 - 2$, so the algebraic degree of $\sqrt{2}$ over $\mathbb{Q}$ is 2
  • The dimension of $\mathbb{Q}(\sqrt{2})$ as a vector space over $\mathbb{Q}$ is also 2, with basis $\{1, \sqrt{2}\}$
• The minimal polynomial of $i + \sqrt{2}$ over $\mathbb{Q}$ is $x^2 - 2x - 3$, so the algebraic degree of $i + \sqrt{2}$ over $\mathbb{Q}$ is 2
  • The dimension of $\mathbb{Q}(i + \sqrt{2})$ as a vector space over $\mathbb{Q}$ is also 2, with basis $\{1, i + \sqrt{2}\}$

Minimal Polynomials vs Field Extensions

Isomorphism and Degree

• If $\alpha$ is algebraic over $F$ with minimal polynomial $m(x)$, then $F(\alpha)$ is isomorphic to the quotient ring $F[x]/(m(x))$
• The degree of the field extension $F(\alpha)$ over $F$ is equal to the algebraic degree of $\alpha$ over $F$
  • This follows from the isomorphism between $F(\alpha)$ 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 $\alpha$ and $\beta$ are algebraic over $F$ with minimal polynomials $m(x)$ and $n(x)$ respectively, then $\alpha + \beta$ and $\alpha\beta$ are also algebraic over $F$
  • The minimal polynomial of $\alpha + \beta$ divides the polynomial $m(x + y)$ in $F[x, y]$
  • The minimal polynomial of $\alpha\beta$ divides the polynomial $m(xy)$ in $F[x, y]$
• If $\alpha$ is algebraic over $F$ and $\beta$ is algebraic over $F(\alpha)$, then $\beta$ is algebraic over $F$
  • The minimal polynomial of $\beta$ over $F$ divides the polynomial obtained by substituting the minimal polynomial of $\alpha$ into the minimal polynomial of $\beta$ over $F(\alpha)$
  • This result allows us to build larger field extensions by adjoining algebraic elements step by step (example: $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ can be constructed by first adjoining $\sqrt{2}$ to $\mathbb{Q}$, then adjoining $\sqrt{3}$ to $\mathbb{Q}(\sqrt{2})$)