User unknown - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:33:52Z http://mathoverflow.net/feeds/user/10189 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43888/local-artin-algebras/43917#43917 Answer by unknown for local Artin algebras unknown 2010-10-28T00:24:47Z 2010-10-28T00:24:47Z <p>Bugs Bunny's answer made me realize that what you are after is true. You have to formulate it more clearly though, I think. Let me then add few things to his answer.</p> <p>Consider the canonical decomposition $A\simeq A_1\times\ldots\times A_n$ of the Artin ring $A$ into the product of local Artin rings $A_i$. The identity $1_A=(1,\ldots, 1)$ can be written as the sum $(1,0,\ldots,0)+\ldots+(0,\ldots,0,1)$, where we call the $i$-th summand $p_i$, following Bags Bunny. We have $p_i^2=p_i$, $p_ip_j=0$ for $i\neq j$, and $1=p_1+\ldots+p_n$. Using these projectors $p_i$ now you can decompose the abelian group $A_0$ underlying the ring $A$ into the sum of the abelian groups $p_iA$: $A_0\simeq p_1A\oplus\ldots\oplus p_nA$. This decomposition has the property that the distinguished element $1_A$ decomposes as $(p_1\cdot 1_A,\ldots,p_n\cdot 1_A)$, moreover $p_iA$ is identified, as $A$-module, to $A_i$ and the distinguished elements $p_i 1_A$ and $1_{A_i}$ correspond to each other under this identification.</p> <p>Now, in what sense this decomposition of $A_0$ is unique?</p> <p>Assume that you are given a decomposition of $A_0\simeq M_1\oplus\ldots\oplus M_n$ into the sum of abelian groups $M_i$ together with the data of isomorphisms of $A$-modules $\varphi_i:A_i\rightarrow M_i$, for all $i$, such that the distinguished element $1_A$ of $A_0$ decomposes as $\varphi_1(1_{A_1})+\ldots+\varphi_n(1_{A_n})$, then the decomoposition $A_0\simeq M_1\oplus\ldots\oplus M_n$ coincide with that described in the first paragraph. </p> http://mathoverflow.net/questions/42853/an-explicit-computation-in-class-field-theory An explicit computation in class field theory unknown 2010-10-20T00:41:02Z 2010-10-21T00:45:26Z <p>Let $K$ be the imaginary quadratic field obtained by joining $\sqrt{-1}$ to the field of rational numbers $Q$. I would like to describe the extension $K^{ab}/Q^{ab}$, where for $F$ a number field, $F^{ab}$ denotes its maximal abelian extension (everything is taking place inside a big fixed field...).</p> <p>More precisely I would like to know the Galois group and the ramification properties of such extension. Is this possible/easy? I suppose one should look at the kernel of the norm map between Idele class groups $N_{K/Q}:I_K\rightarrow I_Q$. But at the moment it is not clear to me how to get the answer. Any hint or comment would be appreciated. Thanks.</p> <p>EDIT: Probably the idele class group of a number field $F$ should be denoted by $J_F$. Or by anything other than $I_F$...</p> http://mathoverflow.net/questions/44018/why-is-symmetric-group-not-matrix Comment by unknown unknown 2010-10-28T21:02:25Z 2010-10-28T21:02:25Z Assume that your infinite symmetric group $G$ has an $n$ dimensional representation over a field $k$. You will be certainly able to find infinitely many elements of $G$ of order two that commute with each other. If $char(k)\neq 2$ then you can simultaneously diagonalized the matrices corresponding to these elements of order two. But then you see that you have only 2^n-1 choices for them. Contradiction. Does it work? http://mathoverflow.net/questions/43888/local-artin-algebras/43904#43904 Comment by unknown unknown 2010-10-28T00:39:17Z 2010-10-28T00:39:17Z Exactly, the question is not clear as it is formulated. I propose below an interpretation of the correct question to ask. http://mathoverflow.net/questions/43888/local-artin-algebras Comment by unknown unknown 2010-10-27T23:20:37Z 2010-10-27T23:20:37Z Let $k$ be an infinite field of characteristic $p&gt;0$. Then there are infinitely many field homomorphisms $\varphi_\alpha:k\rightarrow k$ where $\alpha$ belongs to some infinite indexing set I. Then $k$ admits infinitely many &quot;embedding&quot; in $k\times k$ that split the projection into the first factor: take the maps $x\rightarrow (x, \varphi_\alpha(x))$ for $\alpha\inI$. http://mathoverflow.net/questions/43888/local-artin-algebras Comment by unknown unknown 2010-10-27T22:36:44Z 2010-10-27T22:36:44Z ... the connected components of Spec(A) http://mathoverflow.net/questions/43888/local-artin-algebras Comment by unknown unknown 2010-10-27T22:36:15Z 2010-10-27T22:36:15Z You have to be careful when considering the sum A_1 + ... + A_n. Probably if you are interested in commutative rings then you should call it a product. This would also make clear that the you do not have, in general, maps from A_i to A, but rather from A to A_i (A_i in the way you are looking at it is NOT subring) Considering all of this the answer to your question I think should still be no: if you take a field k and consider k \times k then you would have at least three ways to embed k in k \times k. In more geometric terms the factors A_i, when not further decomposable correspond to http://mathoverflow.net/questions/43095/hilbert-modular-newforms Comment by unknown unknown 2010-10-22T00:13:18Z 2010-10-22T00:13:18Z You can read these things in a paper of Casselman in the second of Antwerp volumes &quot;Modular functions of one variable&quot; (pp 119-120 might be useful). I think that a theorem of the type $a_p\neq 0$ can be proved only if one knows a priori that $\pi_p$ is not supercuspidal. http://mathoverflow.net/questions/43095/hilbert-modular-newforms Comment by unknown unknown 2010-10-22T00:07:13Z 2010-10-22T00:07:13Z In representation theoretic terms if $f$ is a classical newform of conductor $N$ and $\Pi$ is the associated automorphic representation of $GL_2$, then the Hecke eigenvalue $a_p$ is closely related to the local component $\Pi_p$ at $p$ of $\Pi$. If $p$ does not divide $N$, then $\Pi_p$ is obtained by inducing two characters from the Borel that can be determined from $a_p$. If $p$ does divide $N$ then the picture is (I think) more complicated. However if $\Pi_p$ is a twist by a character $\chi$ of the special representation, then the eigenvalue $a_p$ is related to $\chi$ by a simple formula. http://mathoverflow.net/questions/42853/an-explicit-computation-in-class-field-theory/42861#42861 Comment by unknown unknown 2010-10-21T01:10:26Z 2010-10-21T01:10:26Z @Alex: I do not understand your comment. What does the Tate module of some elliptic curve have to do with the group written by Emerton above? http://mathoverflow.net/questions/42853/an-explicit-computation-in-class-field-theory/42861#42861 Comment by unknown unknown 2010-10-21T00:59:30Z 2010-10-21T00:59:30Z @Emerton: Thanks for your answer! http://mathoverflow.net/questions/42853/an-explicit-computation-in-class-field-theory/42854#42854 Comment by unknown unknown 2010-10-20T00:57:30Z 2010-10-20T00:57:30Z Thanks! Can I then answer my question with this information?