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Nov 30, 2010 at 7:17 comment added Marc Nieper-Wißkirchen @Laurent: I like the terminology you suggested although it is not established. Given the discussion above and the fact that no one could provide a usual terminology it seems pretty clear that the answer to my original question is No.
Nov 22, 2010 at 7:45 comment added Laurent Moret-Bailly 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.
Nov 22, 2010 at 3:14 history edited Allen Knutson CC BY-SA 2.5
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Nov 21, 2010 at 1:02 comment added Gerry Myerson 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.
Nov 21, 2010 at 0:54 comment added Gerry Myerson I wouldn't recommend calling it a primitive element, as there is already such a thing as The Primitive Element Theorem, and an element primitive in the sense of that theorem need not be primitive in your sense.
Nov 20, 2010 at 21:14 comment added Marc Nieper-Wißkirchen @Martin: You can find a link to the script of the course on my webpage. During the next weeks I will put further chapters online. As to your other question: I haven't made up my mind yet how I will present Galois theory in characteristic $p$ when more than one non-separable element is adjoined. It has also something to do with that I don't have factorisation methods for non-separable polynomials over non-separable extensions in general. (This is important as I want my reasoning to be constructively valid. See also here: rmmc.asu.edu/TO%20DOUGLAS/RMJ/vol12/vol12-1/min.pdf.)
Nov 20, 2010 at 20:26 comment added Marc Nieper-Wißkirchen @Jim: I agree with you that Galois theory should be treated purely algebraically without constructing the field of complex numbers or proving the so-called fundamental theorem of algebra. However, it is due to didactical reasons that I make use of this theorem as it allows me to develop Galois' theory without having to talk about the abstract concept of a field. Furthermore, I need the complex numbers when talking about the various impossibility theorems of circle and ruler constructions. After the first part of the course I will mention Kronecker's construction of abstract splitting fields.
Nov 20, 2010 at 19:41 comment added Jim Humphreys @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.
Nov 20, 2010 at 19:21 comment added Martin Brandenburg Dear Marc, is there really a need for a special terminology? And how can you develope galois theory without talking about normal extensions which are not given by a primitive element? Anyway, your approach sounds interesting. Please let me know if you write a script for the course.
Nov 20, 2010 at 18:31 comment added Marc Nieper-Wißkirchen @Jim: In the first half of the course, everything is inside $\mathbf C$ so there is no need for abstract field extensions. The Galois group of an algebraic number $\alpha$ over an algebraic number $\beta$ (where $\alpha$ is rational in $\beta$) is being be defined as those permutations of the roots of the minimal polynomial of $\alpha$ over $\beta$ that respect all algebraic relations over $\beta$ between the roots. So I don't have to talk about field automorphisms.
Nov 20, 2010 at 18:25 comment added Marc Nieper-Wißkirchen @Jum: In my case, saying "$\alpha$ is normal" would be an abbreviation for "$alpha$ is a primitive element for itself and all of its conjugates". Actually, this may suggest to call an element $\alpha$ with this property a primitive element.
Nov 20, 2010 at 18:24 comment added Jim Humphreys @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.
Nov 20, 2010 at 18:21 comment added Marc Nieper-Wißkirchen @Pete: I am currently teaching a course on Galois theory. In the first part of this course I don't want to put too much emphasis on field extensions so I am using the more elementary notions of algebraic elements, polynomials and Galois groups of roots of separable polynomials. For example, I would like to be able to say something like: "If x is normal over the rationals, then the order of the Galois group of x and its conjugates equals the degree of x." or "If x is normal and its degree is a power of two over the rationals, then x is constructible by circle and ruler."
Nov 20, 2010 at 18:16 comment added Jim Humphreys I too am unaware of any "established terminology" along this line. Even if you call $\alpha$ normal over $K$ (which may well have occurred in some sources), this doesn't abbreviate the description noticeably. And how many times will the need come up to use such special terminology? Would a reader know immediately what you meant by it?
Nov 20, 2010 at 18:08 history edited Pete L. Clark
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Nov 20, 2010 at 18:06 comment added Pete L. Clark 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.
Nov 20, 2010 at 17:30 history asked Marc Nieper-Wißkirchen CC BY-SA 2.5