Fields & Field Extensions
Fields & Field Extensions
A field $F$ is a commutative ring where every non-zero element is a unit. Field extensions $K/F$ are fields $K$ containing $F$ as a subfield. The degree $[K:F]=\dim_F K$ measures the size of the extension. Simple extensions $F(\alpha)$ are generated by a single element; if $\alpha$ is algebraic over $F$ (satisfies a polynomial with coefficients in $F$), then $F(\alpha)\cong F[x]/\langle m_\alpha\rangle$ where $m_\alpha$ is the minimal polynomial.
Algebraic & Transcendental Extensions
$\alpha\in K$ is algebraic over $F$ if $p(\alpha)=0$ for some non-zero $p\in F[x]$; otherwise transcendental. The minimal polynomial $m_\alpha$ is the monic irreducible polynomial in $F[x]$ with $m_\alpha(\alpha)=0$; $[F(\alpha):F]=\deg m_\alpha$. Tower law: if $L/K/F$ are field extensions, $[L:F]=[L:K][K:F]$. Examples: $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ (minimal poly $x^2-2$); $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$; $\pi$ and $e$ are transcendental over $\mathbb{Q}$.
Splitting Fields
The splitting field of $f\in F[x]$ is the smallest extension $K/F$ over which $f$ factors into linear factors. $K$ is unique up to $F$-isomorphism. Construction: if $f$ has no root in $F$, adjoin a root of an irreducible factor, repeat. Example: splitting field of $x^3-2$ over $\mathbb{Q}$ is $\mathbb{Q}(2^{1/3},\omega)$ where $\omega=e^{2\pi i/3}$; degree $[K:\mathbb{Q}]=6$. Finite fields: $\mathbb{F}_{p^n}$ is the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$; $[\mathbb{F}_{p^n}:\mathbb{F}_p]=n$.
Example 1
Find $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]$ and show $\sqrt{6}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$.
Solution: $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. Is $\sqrt{3}\in\mathbb{Q}(\sqrt{2})$? If $\sqrt{3}=a+b\sqrt{2}$, then $3=a^2+2b^2+2ab\sqrt{2}$, so $ab=0$. If $b=0$: $\sqrt{3}=a\in\mathbb{Q}$, impossible. If $a=0$: $\sqrt{3}=b\sqrt{2}$, so $3/2=b^2\in\mathbb{Q}^2$, impossible. So $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$, and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]=2$, giving $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4$. $\sqrt{6}=\sqrt{2}\cdot\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. ✓
Example 2
Construct $\mathbb{F}_4$ and list its elements.
Solution: $\mathbb{F}_4=\mathbb{F}_2[x]/\langle x^2+x+1\rangle$ ($x^2+x+1$ is irreducible over $\mathbb{F}_2$: no roots $0,1\in\mathbb{F}_2$). Elements: $\{0,1,\alpha,\alpha+1\}$ where $\alpha^2=\alpha+1$ (since $\alpha^2+\alpha+1=0$). Multiplication: $\alpha(\alpha+1)=\alpha^2+\alpha=(\alpha+1)+\alpha=1$ (so $\alpha^{-1}=\alpha+1$); $(\alpha+1)^2=\alpha^2+1=\alpha+1+1=\alpha$. The non-zero elements $\{1,\alpha,\alpha+1\}$ form a cyclic group $\mathbb{Z}_3$ under multiplication ($\alpha$ has order 3).
Practice
- Prove $[K:F]=1$ iff $K=F$.
- Show $\mathbb{Q}(\sqrt[3]{2})$ has no subfields strictly between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{2})$.
- Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$.
- Show that $\mathbb{F}_{p^n}^*$ is cyclic (all finite fields have cyclic multiplicative groups).
Show Answer Key
1. $[K:F]=\dim_F K=1$ iff $K=F\cdot1=F$. Every element of $K$ is a scalar multiple of $1\in F$, so $K=F$.
2. $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ (minimal polynomial $x^3-2$ is irreducible by Eisenstein at $p=2$). Any subfield $L$ with $\mathbb{Q}\subsetneq L\subsetneq\mathbb{Q}(\sqrt[3]{2})$ would have $[L:\mathbb{Q}]|3$, but the only divisors of 3 are 1 and 3. So no proper intermediate field exists.
3. Let $\alpha=\sqrt{2}+\sqrt{3}$. Then $\alpha^2=5+2\sqrt{6}$, so $\alpha^2-5=2\sqrt{6}$, $(\alpha^2-5)^2=24$, $\alpha^4-10\alpha^2+1=0$. The minimal polynomial is $x^4-10x^2+1$ (irreducible over $\mathbb{Q}$ by checking no rational roots and no quadratic factorization).
4. The multiplicative group $\mathbb{F}_{p^n}^*$ has order $p^n-1$. For any finite abelian group, if it's not cyclic, there exist distinct primes $d_1|d_2$ with $x^{d_1}=1$ having more than $d_1$ solutions, contradicting the fact that a polynomial of degree $d$ over a field has at most $d$ roots. So $\mathbb{F}_{p^n}^*$ is cyclic.