LEC ALG3 17

Claim 325.1(Claim).

Let $F,K$ be fields. Let $\{F_i{\}}_{i\in \mathbb{N}}$ be fields such that $F_i\subseteq F_{i+1}$ and $F\subseteq F_i\subseteq K$ for every $i$. Then ${\bigcup }_{i\in \mathbb{N}}{F}_i$ is a subfield of $K$.

9d1148

If $K/F$ is an algebraic extension, is $[K:F]$ always finite? No. Consider

$$\mathbb{Q}\subseteq \mathbb{Q}[\sqrt[2]{2}]\subseteq \mathbb{Q}[\sqrt[4]{2}]\subseteq \mathbb{Q}[\sqrt[8]{2}]\subseteq \cdots \subseteq \mathbb{Q}[\sqrt[2^n]{2}]\subseteq \cdots \subseteq \mathbb{C}.$$

$[\mathbb{Q}[\sqrt[2^i]{2}]:\mathbb{Q}]=2^i$. ${\bigcup }_{i\in \mathbb{N}}\mathbb{Q}[\sqrt[2^i]{2}]$ is a field by Claim 1, and is clearly an algebraic extension of $\mathbb{Q}$. However, $\left[{\bigcup }_{i\in \mathbb{N}}\mathbb{Q}[\sqrt[2^i]{2}]:\mathbb{Q}\right]$ cannot be finite; easy contradiction if it were.

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If $\psi :F[x]\to K$ is a ring homomorphism, then $\ker (\psi )$ contains only one irreducible element up to association ($F[x]$ is a PID!).

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[!Definition]
Let ${\alpha }_1,\dots ,{\alpha }_r\in K$ be algebraic. Then $F({\alpha }_1,\dots ,{\alpha }_n)$ is the smallest subfield of $K$ containing ${\alpha }_1,\dots ,{\alpha }_r$ and $F$.

[!Proposition]
$F\subseteq L\subseteq K$. Let $n=[L:F]<\infty$ and $m=[K:L]<\infty$. Then $[K:F]<\infty$.

[!Proof]-
Let $\{x_1,\dots ,x_n\}$ be a $F$-basis for $L$. Let $\{y_1,\dots ,y_m\}$ be an $L$-basis of $K$.
Claim: $\{x_i{y}_j:1\le i\le n,1\le j\le m\}$ is a spanning set. This is clear, so the lemma follows. However, the set above also happens to be a basis. This is also clear.

[!Lemma]
${\alpha }_1,\dots ,{\alpha }_r\in K\supseteq F$ are algebraic over $F$. Then $F({\alpha }_1,\dots ,{\alpha }_n)$ is a finite extension of $F$, and hence an algebraic extension of $F$.

[!Proof]-
By induction on $n$. $F\subseteq F({\alpha }_1)$ is finite, so $[F({\alpha }_1):F]<\infty$ By induction hypothesis, $[F({\alpha }_1,\dots ,{\alpha }_{n-1}):F]<\infty$. $[F_{n-1}:F]<\infty$. Want $[F_n:F]<\infty$.

Note that $F_n=F_{n-1}({\alpha }_n)$, because $F_{n-1}$ is the smallest subfield of $K$ containing $\alpha ,\dots ,{\alpha }_{n-1}$ and ${\alpha }_n$ is algebraic over $F_{n-1}$. The rest follows from the previous proposition.

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