MATH 3231 Class Notes 02-24-2025

notefield-theoryvector-spaces

Field Extension

Defined Prime Subfield

Defined Morphism of Field Extensions

Defined Degree of Extensions

Theorem

for $K\subseteq E\subseteq F$ extensions, then

$$[F:K]=[E:K]\cdot [F:E].$$

Proof

Assume: $[E:K]=n<\infty$, and $[F:E]=m<\infty$. Assume there also exists a basis $e_1,\dots ,e_n\in E$ over $K$, and $f_1,\dots ,f_m$ a basis over $E$.

Claim: $(e_i\cdot f_j{)}_{\underset{1\le j\le m}{1\le i\le n}}$ is a basis of $F$ over $K$ (this is the product in $F$). Span: Let $f\in F⟹\exists {\epsilon }_1,\dots ,{\epsilon }_m\in E,f={\epsilon }_1{f}_1+\cdots +{\epsilon }_m{f}_m$ in $F$. For all $1\le j\le m$, there exists $c_{ij}\in K$ (for $1\le i\le n$) such that ${\epsilon }_j=c_{1j}{e}_1+\cdots +c_{nj}{e}_n$ in $E$ $⟹$ $f={\sum }_{ij}^{}{c}_{ij}\cdot (e_{ifj})$ in $F$ $⟹$ the $e_{ifj}$ span $F$ over $K$.

Linear independence: Want that if…

Lemma

Take some field $K$, and polynomial $P(x)=x^n+a_{n-1}{x}^{n-1}+\cdots +a_0\in K[x]$. Assume $P(x)$ is irreducible, then,

$${}^{K[x]}{/}_{(P(x))}\text{ is a field and }\left[{}^{K[x]}{/}_{(P(x))}:K\right]=n.$$

Remark

If $n>1$, $P(x)\in K[x]$ monic irreducible of degree $n$ $⟹$ $P(x)$ has no roots in $K$, but $K[x]$ is a field, $\alpha \in K[x]$ is a root of $P(x)$.