22 lines
1.4 KiB
Markdown
22 lines
1.4 KiB
Markdown
# Dummit & Foote Chapter 12 - Field Theory
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## Section 13.1 Basic Theory of Field Extensions
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**Definition**. The *charactaristic* of a field $F$ is the smallest positive integer $p$ such that $1_F * p = 0$. It follows that $p$ is $0$ or prime, and $p \alpha = 0$ for any $\alpha \in F$.
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**Definition**. If $K, F$ are fields such that $F \subseteq K$, then $K$ is an *extension field* or *extension* of $F$, denoted $K / F$.
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**Definition**. The *degree* (or *relative degree* or *index*) of $K/F$, denoted $[K:F]$, is the dimension of $K$ as a $F$-vector space.
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**Theorem**. Let $F$ be a field, $p(x) \in F[x]$. Then, there exists a $K$ such that $p(x)$ has a root in $K$.
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**Theorem**. Let $F$ be a field, $p(x) \in F[x]$. Then, $K = \frac{F[x]}{(p(x))}$ and $\theta = x \amod{p(x)}$, $K$ has a basis of $1, \theta, \ldots, \theta^{n-1}$ where $n = \deg(p)$.
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**Theorem**. Let $K/F$ and $\alpha, \beta, \ldots \in K$. Then, the smallest subfield of $K$ containing $F$ and $\alpha, \beta, \ldots$ is $F(\alpha, \beta, \ldots)$, which is the *field generated by $\alpha, \beta, \ldots$ over $F$*.
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**Definition**. If $K$ is generated by $F(\alpha)$, then $K$ is a *simple extension* of $F$.
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**Theorem**. Let $F$ be a field, $p(x) \in F[x]$ be irreducible. Then, if $\alpha$ is a root of $p(x)$ and $K$ is an extension of $F$ containing $\alpha$, then $F(\alpha) \cong \frac{F[x]}{(p(x))}$.
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TODO
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