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Let $L/K$ be a field extension. Let $\alpha \in L$ be algebraic over $K$. Is there an established terminology for the property of $\alpha$ that $K(\alpha)/K$ is a normal field extension? Would you call $\alpha$ normal over $K$ in case $K(\alpha)/K$ is normal?

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    $\begingroup$ I have been studying field theory for some time, and I have never encountered any terminology for the situation you describe. My first instinct was that calling such an element "normal" is not such a great idea, but I find that I can't defend that at all: upon reflection, it seems reasonable enough. $\endgroup$ Nov 20, 2010 at 18:06
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    $\begingroup$ @Marc: You can't avoid field extensions systematically in such questions, since at some point you have to compare different extensions of a given field which are isomorphic. Elements outside the ground field have to live somewhere definite. $\endgroup$ Nov 20, 2010 at 18:24
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    $\begingroup$ @Marc: Not to prolong the dialogue too much, for me an algebraic subject like Galois theory is better treated algebraically without presupposing knowledge of complex numbers and such. In any case, Galois theory requires systematic study of isomorphisms and automorphisms for finite field extensions: consider the Galois group. Galois himself had the insight for this but lacked the language to make it transparent. $\endgroup$ Nov 20, 2010 at 19:41
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    $\begingroup$ The phrase "normal element" seems to be taken already, too, as it is used in the literature to refer to an element which, with its conjugates, forms a basis (called a "normal basis") for the extension. In that sense, $i$ is not a normal element of ${\bf Q}(i)/{\bf Q}$, since $\lbrace i,-i\rbrace$ is not a basis. $\endgroup$ Nov 21, 2010 at 1:02
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    $\begingroup$ I agree that "normal element" is bad, primarily because "normal extension" isn't that terrific in the first place. "Saturated extension" would be better, perhaps. For an element, if you really need terminology, let me suggest "gregarious" or "fraternal" since it cannot leave its conjugates. $\endgroup$ Nov 22, 2010 at 7:45

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