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September 19

Field extensions

Hi. I'm working through a p-adic analysis book, and there's a quick review of fields at the beginning of chapter 3. I read the following there:

A field extension K of F is an F-vector space; if it is finite-dimensional, it must be an algebraic extension, and its dimension is called the degree [K:F]. If $\alpha \in K$ has the property that every element of K can be written as a rational expression in $\alpha$ , we write $K=F(\alpha )$ and say that K is the extension obtained by adjoining $\alpha$ to F.

Two paragraphs later:

If F is a perfect field, then any finite extension K of F is of the form $K=F(\alpha )$ for some $\alpha \in K$ . $\alpha$ is called a primitive element. Knowing a primitive element $\alpha$ of a field extension K makes it easier to study K, since it means that everything in K is a polynomial in $\alpha$ of degree <n.

Super. So... the first paragraph there says that everything in K is a rational expression in $\alpha$ . The second says that everything in K is a polynomial in $\alpha$ . Those aren't equivalent, are they? What polynomial in $\alpha$ is the same as ${\frac {1}{1-\alpha }}$ , for example? What am I missing? Is the polynomial thing something extra that we get when F is perfect? -GTBacchus^(talk) 00:10, 19 September 2009 (UTC)[reply]

They're equivalent as long as α is algebraic over K. Then any rational expression in α over K can be rewritten as a polynomial, using the extended Euclidean algorithm. For example, if α is a root of the polynomial x²-x+1, then 1/(1-α)=α. Algebraist 00:59, 19 September 2009 (UTC)[reply]

The algorithm at that link seems to apply in a finite field. What if we're talking about an extension of

\mathbb {Q}

, for example? That's a perfect field, because is has characteristic 0, right? -GTBacchus^(talk) 01:02, 19 September 2009 (UTC)[reply]

The algorithm works whenever α is algebraic over the ground field (perfection is not relevant, and neither is finiteness). Algebraist 01:09, 19 September 2009 (UTC)[reply]

Ok, I'll study that. Thank you. -GTBacchus^(talk) 01:29, 19 September 2009 (UTC)[reply]

calculations over three years

I learned one porcelain bowl from the estate of Barbara Hutton was auctioned and sold by Christie's Hong Kong for $22,240,000 in 2006. When I tried to use the inflation calculator provided by the United States Department of Labor to figure out what that sum of money would be today, I ran into a problem. The original sum has to be at least $10,000,000 or less. So if anyone could improvise and help me figure out what $22,240,000 in 2006 would be in today's money, that would be great. Thank you.69.203.157.50 (talk) 05:43, 19 September 2009 (UTC)[reply]

Try the inflation calculator on $22,240 and multiply the result by 1,000. Bo Jacoby (talk) 06:14, 19 September 2009 (UTC).[reply]

Method to generate symmetries of a point lattice

I am interested in algorithmically generating the symmetries of a 2D or 3D point lattice, given reduced lattice basis vectors. I know that there are a finite number of crystallographic point groups in these dimensions, but I am wondering how those tables of symmetries were generated. Is there an automatic way of discovering lattice symmetries? Victor Liu (talk) 19:39, 19 September 2009 (UTC)[reply]

One way you could go about it is to begin at the origin and generate all the lattice points near the origin, expanding your search outward until you have in your collection 3 points that are linearly independent (for a 3D lattice). Then divide all the points in your collection into sets of points that are equidistant from the origin. Any symmetry has to map the points in each set into one another, and any isometry that successfully maps all the points in your collection to other points in the collection is a symmetry of the lattice. That should narrow down the search for what isometries work to something finite. There might be something more clever though. Are you looking to implement this on a computer? Rckrone (talk) 06:56, 20 September 2009 (UTC)[reply]

Yes, I plan to implement this on a computer. I thought it would be interesting to be able to generate the standard tables of crystallographic point groups in 2D and 3D, and then see if I can go to higher dimensions. That's the mathematically curious reason. The practical reason is to implement [1] very generally, among other possible applications. Your proposal sounds good. I'm think to use the set of equidistant nearest neighbors from the origin, and trying rotations and flips on them. I still need to convince myself this works. 76.102.161.22 (talk) 08:36, 20 September 2009 (UTC)[reply]

made link visible - no reference section on this page83.100.251.196 (talk) 09:47, 20 September 2009 (UTC)[reply]

Just the nearest neighbors won't be enough in general, since that set might not span the lattice. Consider for example the 3D lattice Z×2Z×3Z. You may need to look at the sets of points at several different distances so that the union of the sets spans. For the nD case, you could need up to n different sets. Rckrone (talk) 23:18, 20 September 2009 (UTC)[reply]