yeah, I know about Fourier series representations of signals and such, but I am trying to explain this to my non-mathematical friend, and I don't want to overwhelm him with these kinds of technical things. Anything lower-tech?
So, for instance, I was able to impress upon him that he thinks in high-dimensional space all the time by appealing to finance: the market can be viewed as a 1d curve sitting in R^n where n is the number of commodities under consideration
You have an infinite hallway in Hilbert's hotel. In each room there are various things like furniture etc. that is worth money. The hotel staff moves everything on one side of the hallway to the other side, thus increasing the value of the rooms in the latter. We can view this as an infinite-dimensional array of values - two of them that we are adding together.
I suppose it's also not too much of a stretch to imagine a market with infinitely many commodities, once you've convinced yourself it works for huge values of n
Congratulations, anon. I think I just discovered something too. Could you confirm that in $\mathbb F[x]/(x^n)$ the units are exactly the elements $a_0+a_1x^2+\ldots+a_{n-1}x^{n-1}+(x^n)$ such that $a_0\neq 0$? I think I found a nice way of proving this.
I wouldn't have thought it was possible to be too loose with these words, but never mind. Continue your Galois Theory conversation - I'll try to grok it from the transcript.
In this context $f_{1}$ and $f_{2}$ are separable with splitting fields $K_{1}$ and $K_{2}$ so, the compositum $K_{1}K_{2}$ is the splitting field of the squarefree part of $f_{1}(x)f_{2}(x)$
Can we make an answer community wiki, get votes on it, then change it back to normal and get the rep? I have a partial answer on someone's question and plan to ask my own question in order to finish it..
So if we write $A_3$ as $\{0,a,b\}$, then the four subgroups of order 3 of $A_3\times A_3$ would be $\{(0,0),(a,0),(b,0)\}, \{(0,0),(a,a),(b,b)\}, \{(0,0),(a,b),(b,a)\}$ and $\{(0,0),(0,a),(0,b)\}$ with all the usual operations.
Determine the Galois group of $(x^{3}-2)(x^{3}-3)$ over $\Bbb{Q}$. Determine all the subfields which contain $\Bbb{Q}(\rho)$ where $\rho$ is a primitive $3^{\text{rd}}$ root of unity.
But if the Galois group is isomorphic to $A_{3}\times A_{3}$, then its degree is 9, but then this thread implies that $\rho$ is in $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$.
So either, $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ is not a subfield of $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3},\rho)$ or the Galois group is not $A_{3}\times A_{3}$.
I don't know. It must be right? But the tower law requires that $[\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3‌​},\rho):\Bbb{Q}]=[\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3‌​},\rho):\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})][\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3‌​}): \Bbb{ Q }]$.
Yeah, I'm starting to think that must be the case. But then I'm not sure where I went wrong in applying the theorem I used to get that conclusion. On another note. What do you think $\Bbb{Q}(\sqrt[3]{2},\rho)\cap\Bbb{Q}(\sqrt[3]{3},\rho)=$ ?
Well $\rho$ is a $3^{\text{rd}}$ root of unity, so the degree of it's minimal polynomial is $\varphi(3)$ where $\varphi$ is the Euler $\varphi$ function
So I know that the Galois group is isomorphic to the direct product $\operatorname{Gal}(\Bbb{Q}(\sqrt[3]{2},\rho)/\Bbb{ Q }(\rho))\times\operatorname{Gal}(\Bbb{ Q }(\sqrt[3]{3},\rho)/\Bbb{Q}(\rho))$
Hmm. When posting a question, the system seems to automatically permute the order of the tags you entered into something different. How does that work?
I never know how to answer that question, because I haven't taken classes outside of high school (vector calc, lin alg and diff equ). I've learned a little bit of algebra, number theory etc. reading for fun. (I have a collection of notes and texts on my hard drive.)
I'm really unbalanced - I know a lot more from the analytic side than the algebraic side. But I want to work up to Langland's program. (Also another interest might be representation theory, but I haven't really looked at it deeply.)
I guess that's not really something too "particular." :D
Haha, I think everyone would like to learn a little about the Langland's program. Yeah, if you're interested in that you will HAVE to learn a LOT of representation theory. My (infinitesimal) understanding of Langland's is that it's a correspondence between "algebraic" and "analytic" representations.
Right now? I'm taking a course in homological algebra/group cohomology and a course in "advanced" galois theory (not that advanced). Next term I'm going to be doing algebraic number theory and a very exciting course on the algebraic side of Riemann surfaces.
This will probably just sound really stupid, but I'm too curious not to ask. We have like four guys in our department from Argentina, none are from the same region/school, etc. They all do the same kind of PDE theoretic functional analysis. Is that big in Argentina?
