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Revision difference : field adjunction |
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Version 8 |
| Let $K$ be a field and $E$ an extension field. \,If $\alpha \in E$, then the smallest subfield of $E$, that contains $K$ and $\alpha$, is denoted by $K(\alpha)$. \,We say that $K(\alpha)$ is obtained from the field $K$ by adjoining the element $\alpha$ to $K$ via {\em field adjunction}. |
Let $K$ be a field and $E$ an extension field. \,If $\alpha \in E$, then the smallest subfield of $E$, that contains $K$ and $\alpha$, is denoted by $K(\alpha)$. \,We say that $K(\alpha)$ is obtained from the field $K$ by adjoining the element $\alpha$ to $K$ via {\em field adjunction}. |
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| \begin{thmplain} |
\textbf{Theorem.} \,$K(\alpha)$ is identical with the quotient field $Q$ of $K[\alpha]$. |
| \, $K(\alpha)$ is identical with the quotient field $Q$ of $K[\alpha]$. |
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| \end{thmplain} |
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| {\em Proof.} (1) Because $K[\alpha]$ is an integral domain (as a subring of the field $E$), all possible quotients of the elements of $K[\alpha]$ belong to $E$. So we have |
{\em Proof.} (1) Because $K[\alpha]$ is an integral domain (as a subring of the field $E$), all possible quotients of the elements of $K[\alpha]$ belong to $E$. So we have |
| $$K\cup\{\alpha\} \subseteq K[\alpha] \subseteq Q \subseteq E,$$ |
$$K\cup\{\alpha\} \subseteq K[\alpha] \subseteq Q \subseteq E,$$ |
| and because $K(\alpha)$ was the smallest, then \,$K(\alpha) \subseteq Q.$ |
and because $K(\alpha)$ was the smallest, then \,$K(\alpha) \subseteq Q.$ |
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(2) $K(\alpha)$ is a subring of $E$ containing $K$ and $\alpha$, therefore also the whole ring $K[\alpha]$, that is, \,$K[\alpha] \subseteq K(\alpha)$. \,And because $K(\alpha)$ is a field, it must contain all possible quotients of the elements of $K[\alpha]$, i.e., \,$Q \subseteq K(\alpha)$.
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(2) $K(\alpha)$ is a subring of $E$ containing $K$ and $\alpha$, therefore also the whole ring $K[\alpha]$, that is, \,$K[\alpha] \subseteq K(\alpha)$. \,And because $K(\alpha)$ is a field, it must contain all possible quotients of the elements of $K[\alpha]$, i.e., $Q \subseteq K(\alpha)$.
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