User hagen knaf - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:25:48Z http://mathoverflow.net/feeds/user/1886 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12394/representability-of-finite-metric-spaces/12403#12403 Answer by Hagen Knaf for Representability of finite metric spaces Hagen Knaf 2010-01-20T11:03:34Z 2010-01-20T11:03:34Z <p>The problem under which conditions a finite metric space $(X,d)$ can be embedded isometrically into Euclidean space $\mathbb{R}^n$ is answered in the article</p> <p>C.L. Morgan, EMBEDDING METRIC SPACES IN EUCLIDEAN SPACE, Journal of Geometry. Vol. 5/1 1974</p> <p>The answer is too technical to write it down here, but if I remember it well, the article can be found in the Web.</p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/12309#12309 Answer by Hagen Knaf for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Hagen Knaf 2010-01-19T13:55:41Z 2010-01-19T13:55:41Z <p>The following is about an application of Krull valuations in non-noetherian ring theory:</p> <p>A noetherian normal domain $R$ can be written as an intersection of discrete valuation domains. Moreover these valuation domains are localizations of $R$ - at the prime ideals of height $1$.</p> <p>Of course every normal domain $R$ can be written as an intersection of valuation domains of its field of fractions $K$. Motivated by the situation in the noetherian case one can however try to find families $F$ of valuation domains of $K$ such that $R$ is the intersection of the members of $F$ and $F$ has one or more of the following properties:</p> <ol> <li>All members of $F$ are localizations of $R$ at certain primes (or weaker: for every member of $F$ the values of the elements $r\in R$ are precisely all non-negative values in the value group).</li> <li>All members of $F$ satisfy certain requirements concerning the value groups or the Krull dimension.</li> <li>Every non-zero $r\in R$ lies in the maximal ideal of at most finitely many elements of $F$.</li> </ol> <p>These properties can be used to classify normal domains to a certain extent. In the period roughly between 1950 and 1970 a bunch of articles were published that seemed to follow this idea. Some author names: Paolo Ribenboim, Jim Brewer, Malcolm Griffin.</p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/12297#12297 Answer by Hagen Knaf for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Hagen Knaf 2010-01-19T10:56:14Z 2010-01-19T10:56:14Z <p>Let $F/K$ be an algebraic function field. The set $Z$ of all valuation rings of $F$ that contain $K$ can be identified with the projective limit of all projective (proper) $K$-varieties $X$ having $F$ as their function field. In general $Z$ is a ringed space but not a scheme. By construction it is clear that $Z$ encodes geometric information.</p> http://mathoverflow.net/questions/8324/what-does-linearly-disjoint-mean-for-abstract-field-extensions/8337#8337 Answer by Hagen Knaf for What does "linearly disjoint" mean for abstract field extensions? Hagen Knaf 2009-12-09T11:05:59Z 2009-12-09T11:05:59Z <p>Linear disjointness and its relation to tensor products is explained in detail in Zariski+Samuel, Commutative Algebra - I forgot which volume of the two.</p> <p>There linear disjointness over $k$ of two $k$-algebras $A,B$ is defined only for algebras that are contained in some larger ring $C$.</p> <p>The tensor product is introduced as follows: a product of $A$ and $B$ is a $k$-algebra $C$ and $k$-algebra morphisms $f:A\rightarrow C$, $g:B\rightarrow C$ such that the smallest subalgebra of $C$ containing $f(A),g(B)$ is $C$ itself.</p> <p>A product $C$ of $A$ and $B$ is called tensor product, if $f(A)$ and $g(B)$ are linearly disjoint over $k$.</p> <p>As for the case of two field extensions $E,F$ of $k$ one of which is algebraic over $k$ ($F$ say) one has a surjective map $E\otimes_k F\rightarrow E.F$, where $E.F$ denotes the smallest subring of the algebraic closure of $E$ that contains $E$ and $F$.</p> <p>Since $F/k$ is algebraic, $E.F$ also is the smallest subfield that contains $E$ and $F$.</p> <p>We get the equivalent statements:</p> <p>(1) $E$ and $F$ are linearly disjoint over $k$ within the algebraic closure of $E$.</p> <p>(2) the tensor product $E\otimes_k F$ is a field.</p> http://mathoverflow.net/questions/7469/is-relatively-algebraically-closed-stable-under-finite-field-extensions/7568#7568 Answer by Hagen Knaf for Is (relatively) algebraically closed stable under finite field extensions? Hagen Knaf 2009-12-02T08:51:57Z 2009-12-02T08:51:57Z <p>Counterexample:</p> <p>let $F$ be a non-perfect field of characteristic $p$.</p> <p>Let $L$ be an extension of $F$ of degree $p^2$ such that $L=F(a,b)$ with $a^p,b^p\in F$.</p> <p>The polynomial $f(Y):=Y^p-(a^px^p+b^p)\in F(x)[Y]$ is irreducible, where $F(x)$ is the rational function field in the variable $x$.</p> <p>Consider $F^\prime := F(x,y)$, where $y$ is a root of $f$.</p> <p>Then $F$ is algebraically closed in $F^\prime$: let $K$ be the algebraic closure of $F$ in $F^\prime$. Then $[K:F]=[K(x):F(x)]\leq [F^\prime :F(x)]=p$. Hence $K\neq F$ implies $F^\prime =K(x)$ and thus $y=g(x)\in K[x]$ with $[K:F]=p$ -- in contradiction to the choice of $y$.</p> <p>The tensor product $F^\prime\otimes_F L$ is not a field: the tensor product $F^\prime\otimes_K L$ equals $L(x)[Y]/(f)$. However $f$ is a $p$-th power in $L(x)[Y]$.</p> <p>H</p> http://mathoverflow.net/questions/44/weil-divisors-on-non-noetherian-schemes/6881#6881 Answer by Hagen Knaf for Weil divisors on non Noetherian schemes Hagen Knaf 2009-11-26T09:15:15Z 2009-11-26T09:15:15Z <p>Let R be the integral closure of $\mathbb{Z}$ in the algebraic closure of $\mathbb{Q}$. Then $\dim (R)=1$ and there are infinitely many prime ideals lying above a prime ideal $p\mathbb{Z}$. Hence the closed set $V(pR)$ has the required property.</p> http://mathoverflow.net/questions/5936/whats-so-great-about-blackboards/5950#5950 Answer by Hagen Knaf for What's so great about blackboards? Hagen Knaf 2009-11-18T11:05:10Z 2009-11-18T11:05:10Z <p>The fact that giving a presentation at the black board slows down the speed of presentation and helps the audience to digest the stuff is related to the fact that mathematicians use a language that has a high information density (formulas, diagrams) compared to other fields. Moreover the symbols and other structures used in that language CAN be written on a black board. This is different for a biologist say: using a black board for him/her can be very inefficient, because of the necessity to write down more or less complete sentences or long phrases. This is too much of slowing down. Information dense structures used in biology are pictures or visualizations of numerical data that are hard to create at the black board.</p> http://mathoverflow.net/questions/13016/generic-fiber-of-morphism-between-non-singular-curves Comment by Hagen Knaf Hagen Knaf 2010-01-26T10:04:45Z 2010-01-26T10:04:45Z You need an algebraically closed base field in the statement. Otherwise there can be points $Q\in C_1$ such that the degree $[k(Q):k(\phi (Q))]&gt;1$ in which case the number of points in the fibre $\phi^{-1}(\phi (Q))$ is less than the degree.