Field Theory | Mathematics - Wyatt's Notes

A field extension is an inclusion $F \subseteq E$ of fields. We write $E/F$ and call $E$ an extension field of $F$.

The degree of the extension, denoted $[E : F]$Is the dimension of $E$ as a vector space over $F$.

Proposition 12.1. If $F \subseteq E \subseteq K$ are field extensions, then $[K : F] = [K : E][E : F]$.

Proof. If $\{\alpha_i\}$ is a basis for $E/F$ and $\{\beta_j\}$ is a basis for $K/E$ Then $\{\alpha_i \beta_j\}$ is a basis for $K/F$. Count dimensions. $\blacksquare$

An element $\alpha \in E$ is algebraic over $F$ if there exists a non-zero polynomial $f \in F[x]$ With $f(\alpha) = 0$. Otherwise $\alpha$ is transcendental over $F$.

The minimal polynomial of $\alpha$ over $F$ is the monic polynomial of smallest degree in $F[x]$ Having $\alpha$ as a root.

Proposition 12.2. The minimal polynomial of $\alpha$ over $F$ is irreducible in $F[x]$.

Proof. If $m_\alpha = fg$ with $\deg(f), \deg(g) \lt \deg(m_\alpha)$ Then $f(\alpha)g(\alpha) = 0$ So either $f(\alpha) = 0$ or $g(\alpha) = 0$Contradicting the minimality of $\deg(m_\alpha)$. $\blacksquare$

Theorem 12.3. $\alpha$ is algebraic over $F$ if and only if $[F(\alpha) : F] \lt \infty$. In this case, $[F(\alpha) : F] = \deg(m_\alpha)$.

Proof. If $\alpha$ is algebraic with minimal polynomial $m_\alpha$ of degree $n$ Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $F(\alpha)/F$ (every element can be Reduced modulo $m_\alpha$), so $[F(\alpha) : F] = n$. Conversely, if $[F(\alpha) : F] = n \lt \infty$ Then $\{1, \alpha, \ldots, \alpha^n\}$ is linearly dependent, giving a polynomial relation $f(\alpha) = 0$. $\blacksquare$

Theorem 12.4 (Kronecker”s Theorem). If $F$ is a field and $f \in F[x]$ is irreducible, then $E = F[x] / (f)$ is a field extension of $F$ containing a root of $f$.

Proof. Since $f$ is irreducible and $F[x]$ is a PID, $(f)$ is a maximal ideal, so $E = F[x]/(f)$ Is a field. The element $\alpha = x + (f) \in E$ satisfies $f(\alpha) = f(x + (f)) = f(x) + (f) = (f) = 0$ I.e., $\alpha$ is a root of $f$. $\blacksquare$

Theorem 12.5. For every prime $p$ and every $n \geq 1$There exists a field of order $p^n$ Unique up to isomorphism.

Proof (existence). Consider the splitting field of $f(x) = x^{p^n} - x$ over $\mathbb{F}_p$. The set of roots of $f$ in the splitting field forms a field (since roots are closed under addition, Multiplication, and taking inverses), and it has exactly $p^n$ elements. $\blacksquare$

Proposition 12.6. The multiplicative group $\mathbb{F}_{p^n}^*$ of a finite field is cyclic.

Proof. $\mathbb{F}_{p^n}^*$ is a finite abelian group of order $p^n - 1$. Let $m$ be the largest order of any element. By Lagrange, every element’s order divides $m$. So $x^m = 1$ for all $x \in \mathbb{F}_{p^n}^*$Meaning every element is a root of $x^m - 1$. Since $x^m - 1$ has at most $m$ roots in a field, $p^n - 1 \leq m$. But $m$ divides $p^n - 1$ So $m = p^n - 1$. $\blacksquare$

A field $F$ is algebraically closed if every non-constant polynomial in $F[x]$ has a root in $F$.

Theorem 12.7 (Fundamental Theorem of Algebra). $\mathbb{C}$ is algebraically closed.

Remark. Every field $F$ has an algebraic closure $\overline{F}$: an algebraically closed field That is an algebraic extension of $F$. The algebraic closure is unique up to $F$-isomorphism. For example, $\overline{\mathbb{Q}}$ is the field of all algebraic numbers. It is countable and Infinite-dimensional over $\mathbb{Q}$.

Problem. Compute $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}]$ and find the minimal polynomial of $\sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$.

Problem. Show that $\mathbb{Q}(\sqrt[4]{2})$ is not a Galois extension of $\mathbb{Q}$.

Problem. Construct $\mathbb{F}_9$ as a quotient of $\mathbb{F}_3[x]$.

A field extension answers the question: what happens when you adjoin a new number that was not previously available? Starting from the rationals $\mathbb{Q}$ and adjoining $\sqrt{2}$ creates a larger field $\mathbb{Q}(\sqrt{2})$ that contains all numbers of the form $a + b\sqrt{2}$. The degree of the extension measures how many new independent directions this added element creates. The tower law $[K:F] = [K:E][E:F]$ says that building an extension in stages multiplies the dimensions, just as building a vector space by adding basis vectors multiplies the total dimension.

The minimal polynomial is the irreducible polynomial that the new element satisfies, and its degree equals the extension degree. This connects field theory to polynomial algebra: irreducible polynomials are the “prime building blocks” for constructing new fields, much as prime numbers are building blocks for the integers. Finite fields exist only in sizes $p^n$ and are unique up to isomorphism, which is remarkable: it means $\mathbb{F}_9$ is essentially $\mathbb{F}_3[x]/(x^2+1)$ regardless of which irreducible quadratic you choose. The multiplicative group of any finite field is cyclic, which is why primitive elements exist and why discrete logarithms are well-defined in finite fields.

Theorem 12.8 (Primitive Element Theorem). Every finite separable extension $E/F$ is simple: There exists $\theta \in E$ such that $E = F(\theta)$.

Proof (sketch). If $F$ is infinite, it suffices to find $\theta = \alpha + c\beta$ for suitable $c \in F$ When $E = F(\alpha, \beta)$. Only finitely many values of $c$ fail to work. For $F$ of characteristic $0$Every finite extension is separable, so every finite extension of $\mathbb{Q}$ is simple. $\blacksquare$

Corollary 12.9. Every finite extension of $\mathbb{Q}$ is simple.

Example. $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$.

Mistake 1: Confusing extension degree with polynomial degree The degree $[E : F]$ of a field extension equals the degree of the minimal polynomial of a primitive element only when the extension is simple and generated by that element. For example, $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 4$, not $2$, even though $\sqrt{2}$ and $\sqrt{3}$ each satisfy a quadratic polynomial. The tower law $[K:F] = [K:E][E:F]$ must be applied correctly.

Mistake 2: Assuming algebraic elements generate finite extensions An element $\alpha$ is algebraic over $F$ if and only if $[F(\alpha) : F] < \infty$. This is Theorem 12.3, but students often confuse algebraic with transcendental. An element is transcendental if no polynomial relation exists, meaning $[F(\alpha) : F] = \infty$. For example, $\pi$ is transcendental over $\mathbb{Q}$, so $[\mathbb{Q}(\pi) : \mathbb{Q}]$ is infinite.

Mistake 3: Forgetting that finite fields have characteristic $p$ Every finite field has order $p^n$ for some prime $p$ and integer $n \geq 1$. Students sometimes attempt to construct a finite field of order $6$ or $10$, which is impossible since these are not prime powers. The characteristic of any finite field is prime, and the multiplicative group is cyclic.

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• Rings: Fields are special commutative rings where every nonzero element is a unit.

• Ideals and Quotient Rings: Quotienting a polynomial ring by an irreducible ideal yields a field, the fundamental construction of field extensions.

• Number Theory: Number theory uses finite fields and field extensions in the study of Diophantine equations and cryptography.

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