@Anthony Well, you don't need much. You can easily prove that $2^n > n$, for example, and that gives you $\sqrt[n]{n} < 2$ (but that is not enough, you need more), showing e.g. $n < \left(\frac{3}{2}\right)^n$ for all large enough $n$ however would be enough here.
@Anthony Well, that holds if $n < \left(\frac{3}{2}\right)^n$. So if you can show that for any $n \geqslant 2$, induction gives you the desired inequality for all larger $n$.
@AlexanderGruber Note for example, that left $A$-modules are $\Gamma$-groups where $\Gamma$ is the set of endomorphisms of "multiply to the left", and the Frattini is the intersection of all maximal submodules.
@AlexanderGruber It can be shown that if ${\mathfrak R}(A)$ is the Jacobson radical of $A$, and $M$ is an $A$-module (left), then ${\mathfrak R}(A)M\subseteq {\rm Fr}(M)$
The proof is done by using the Jacobson radical kills simple $A$-modules.
@AlexanderGruber This is basically the observation of Nakayama: if $N$ is maximal, then $M/N$ is simple, so ${\mathfrak R}(A)M/N=\frac{{\mathfrak R}(A)M+N}N=0$ so ${\mathfrak R}(A)M\subseteq N$. This means ${\mathfrak R}(A)M\subseteq{\rm Fr}(M)$.
Thus, if $\mathfrak a$ is an ideal contained in the Jacobson radical and $M={\mathfrak a}M$, then $M=0$; since elements of ${\mathfrak a}M$ are non generators.
Hi everyone. I have this question that I posted on math.SE, and it hasn't received any answers: math.stackexchange.com/q/704758/13524 I'm thinking it would be appropriate to move this question to Math Overflow, since it came up in the course of research.
Is there a way I can ask a moderator to migrate the question? Can I just flag it, or is there some chat room I can ask in?
This doesn't give the stronger that if $M$ is finitely generated and if $\mathfrak a$ is any ideal with $M=\mathfrak a M$; then there is $x=1\mod \mathfrak a$ with $xM=0$, but it is enough to prove the "Nakayama lemma" as stated in my last comment.
@AlexanderGruber I thought I had come up with an answer, but it turned out my answer was incorrect. I'm not sure if I should leave it as an answer, or delete the answer and incorporate it into the question.
Well, I guess it is merely a matter of convention that $\bigcap\varnothing=M$, so the problem may actually be that your proof doesn't adequately handle the case where there are no maximal $\Gamma$-subgroups.
Suppose $M$ is an $A$-module with no maximal submodules. It is clear the set of nongenerators is a subset of $M$. Now, pick an element $x\in M$, and suppose it is not a nongeneator. So there exists $Y\subseteq M$; $x\notin Y$ so that $\langle Y\rangle\subsetneq \langle x,Y\rangle =M$.
Let $\mathscr C$ be the collection of submodules $N$ of $M$ with the property that $Y\subseteq N$, and $x\notin N$.
This is nonempty, and we may order it by inclusion.
If we have a chain in $\mathscr C$, the union contains $Y$, doesn't contain $x$.
It is a submodule, since we have a chain.
So Zorn gives a maximal element, which has to be a maximal submodule, contra the fact there are none.
Since ${\mathfrak R}(A)$ annhilates all simple $A$-modules, ${\mathfrak R}(A)M\subseteq {\rm Fr}(M)$. This means that if $\mathfrak a$ is in the Jacobson radical, $\mathfrak a M$ consists of nongenerators.
@Karl There was another fkaw in my argument but I have corrected it. If M = aM, we cannot immediately conclude M=0 since.the nongenerator defn.allows us to remove an elt when the Y is nonempty. But we can reduce our set of fin many gens to one, hence N is cyclic. Un fact, it is simple, since we may choose any element as a gen, just add it and remove the other. Since a is an ideal inside the Jacobson radical, this kills M, so M=aM=0
Ib particular, since the rationals as a Z module admit no maximal subgroups, every finitely generated subgroup is cyclic by the above
@nerdy Yes, and you can prove it! Two steps: prove that the equivalence class of $e$ is a normal subgroup, and then recall the first isomorphism theorem.
how do we prove that for any congruence relation ~ on (G,*), the equivalence class of e is a normal subgroup ? I'm kinda lost, we didn't define what the congruence relation is yet
If I'm given two regions, could someone give me a tip for finding a bijective map from one to the other?
One of the regions is just the unit-sphere (the domain of the map) and the other is... well... I don't even know what it is: (7x - 3y - z)^2 + (-3x + 7y -z)^2 + (-x -y +3z)^2 <= 100
I have kind-of an idea of what to do, and I've solved problems like this before, but that last shape is blindsiding me. I'm not sure how to handle it in the slightest
I consider something like... I know that the map (call it T) will take boundary points to boundary points, so it must take things of the form "u^2 + v^2 + w^2 = 1" to things of the form... well... whatever that other thing is
The reason I don't ask on the main-site is because someone asked this maybe 10 hours ago and no one answered, so I figure I'd try my luck here. :)
I'm not actually sure what quadratic forms are, either, haha. Are these things typical of a Vector Calculus course? (I haven't heard them in class; but I am interested in figuring out what they are)
What I've been trying to do is express the ellipsoid in spherical coordinates, solve for the coefficients, and basically create a map from the unit sphere in spherical coordinates to the ellipsoid by inspection (which should just consist of tacking on coefficients, right?)
divide by 100, shift the ellipsoid so that it is centered at the origin. write it in the form $X^{T}AX = 1$, orthogonally diagonalize A to that $A = P^{T}DP$, substitute it back into the quadratic form
Let $p$ be a prime and let all variables be in $\Bbb{Z}_p$.
Then you can write the result of
if(i > 0)
k = (a + b)c;
(C code)
as a polynomial $k := i^{p-1} (a+b)c + (1 - i^{p-1}) k$ (notice $:=$ and not $=$). But what about
if (i > j)
k = (a + b)c;
?
If you try converting $...
It's Easter time, and in my workplace we have a "Count the eggs in the jar!" kind of game.
What would be the best mathematical strategy to get as close as possible to the correct count?
Let $A$ be symmetric matrix such that $\exists \lambda_0$ with $\lambda>\lambda_0>0$ for any eigenvalue $\lambda$. Then it is trivial to see that
a) $A$ is positive definite and
b) $\left\|A^{-1}x\right\|\leq \|x\|\lambda_0$.
Any hint why this is trivial to see?
@GabrielR. I'm think about differentiating Fourier sums then setting x to be a certain value, so I think that's the type of convergence that would be useful to me.
@Sawarnik I'm going to be quite busy today :( I've got plans to go and watch a movie and I've gotta downgrade my phone back to Gingerbread because the new firmware sucks.
@DanielFischer Yay, awesome. Ok, here goes: The following is a problem from the Scottish book. The question is, whether a matrix is normal if and only if it is "finite in each row and invertible (in a one-to-one way)"
@Daniel and I'm not clear what it means. Obviously every $n$ by $m$ matrix is finite in each row. So I'm wondering if they are using matrix to mean a linear operator.
Yes, so if that is what is meant by normal, you and I know that a normal matrix has no reason to be invertible. But that would make the question unworthy of a prize of a bottle of wine, so...
@Alyosha you got pointwise convergence of the term-by-term derivative to $f'$ when 1)f is piecewise $C^1$ and regularized 2)$\sum c_n(f) e^{inx} $converges uniformly on any closed interval
@DanielFischer Actually, the original reads "Is a matrix, finite in each row and invertible (in a one-to-one way), equivalent to a normal matrix?" but that's the same as what I translated it to mean, right?
Let $p$ be a prime and let all variables be in $\Bbb{Z}_p$.
Then you can write the result of
if(i > 0)
k = (a + b)c;
(C code)
as a polynomial $k := i^{p-1} (a+b)c + (1 - i^{p-1}) k$ (notice $:=$ and not $=$). But what about
if (i > j)
k = (a + b)c;
?
If you try converting $...
It's Easter time, and in my workplace we have a "Count the eggs in the jar!" kind of game.
What would be the best mathematical strategy to get as close as possible to the correct count?
Let $A$ be symmetric matrix such that $\exists \lambda_0$ with $\lambda>\lambda_0>0$ for any eigenvalue $\lambda$. Then it is trivial to see that
a) $A$ is positive definite and
b) $\left\|A^{-1}x\right\|\leq \|x\|\lambda_0$.
Any hint why this is trivial to see?
