@LeakyNun Sorry for the delay. I decided shortly after making my post that I would try and sleep on it. I don't actually know how to approach either implication (polynomials whose coefficients are themselves polynomials always intimidate me)
For anyone interested, I'm trying to tackle this right now: "Let $F\subseteq K$ be fields. Show that the domain $F+xK[x]$ is integrally closed if and only if $F$ is algebraically closed in $K$"
Suppose that $\mu$ is the measure on $(\mathbb{Z}^+,2^{\mathbb{Z}^+})$ defined by $$\mu(E)=\sum_{n\in\,E}\frac{1}{2^n}$$ Prove that for every $\epsilon>0$, there exists a set $E\subset\mathbb{Z}^+$ with $\mu(\mathbb{Z}^+\setminus\,E)<\epsilon$ such that $f_1,f_2,\dots$ converges uniformly on $E$ for every sequence of functions $f_1,f_2,\dots$ from $\mathbb{Z}^+$ to $\mathbb{R}$ that converges pointwise on $\mathbb{Z}^+$.
I actually had relatively little cheating with my students, because I taught more upper-level courses and hard Honors courses. I caught some cheating on homework, and prosecuted.
LOL, Thorgott. I actually caught some of my own differential geometry students posting their homework on here. A speech in class followed the next day.
Of course. You'll notice that I engage in a lot of socratic conversations and try to get students to figure things out with my hints and responses. But typically someone interrupts by posting a complete solution. That makes me furious.
We typically engage in "office hours discussions" here rather than giving complete solutions, although Mathein and others have been known to do so.
The notion that grades should be based totally on a few exams insults my notion of education. So, yes, I continue to think that graded homework is essential.
Typically. In my advanced courses, I ultimately agreed to give B's if the student attended every lecture, but to earn an A one had to do a certain portion of the homework problems.
Yeah, I just have a slight fear that I might not understand the matertial as well as I would like to think I do. Also, direct in-person interaction is stressful for me
Well, in all seriousness, you should work on that. I presume you will need to do some teaching, and you need to learn to feel comfortable in that venue.
@Rithaniel, I know it's difficult to get over that hurdle, but whether you continue in academia or get a real-world job, you'll have to interact with people.
I think I got my problem. since $\mu(Z^+)=1$, as a single integer can be a subset of $Z^+$, there exists such integer $N$ such that $\mu({N})<\epsilon$. Let $E$ be a set such that its maximum less than $N$
Actually, I have to help grading an exam tomorrow. I've graded assignments before, but not exams. Is there any advice you think is important? @TedShifrin
@Thorgott: Your professor needs to give guidelines and advice. I can't do that in the abstract. Be sure to grade one problem at a time, so that your grading is consistent across the different students. Ideally, come up with a grading rubrik before your start, and modify it slightly (if the professor doesn't give it to you) as necessary. But then that means going back and re-grading to be consistent.
So, on the problem I posted above, I'm taking steps and trying to figure out where to go next
Heya Erico
What I have so far is "Suppose $F+xK[x]$ is integrally closed. Then let $t^{n+1}+\sum_{i=0}^nf_i(x)t^i$ be a monic polynomial, where $f_i(x)\in F+xK[x]$. Then there exists $g(x)\in F+xK[x]$ where $g(x)^{n+1}+\sum_{i=0}^nf_i(x)g(x)^i=0$. Next, we can pull the constant term out of each polynomial in this expression to attain $f+g'(x)^{n+1}+\sum_{i=0}^nf_i'(x)g'(x)^i=0$ where $f\in F$ and $g'(x),f_i'(x)\in xK[x]$"
How I go from here to claiming that $F$ is algebraically closed over $K$ is still unclear to me
Now, we have $\mu(Z^+\setminus\,E)<\epsilon$. Because $\{f_n\}$ converges pointwise on $Z^+$. we can take $\epsilon'=\max\{\epsilon_1,\epsilon_2,\dots,\}$ for all $x\in\,E$ since $E$ is finite. Then $\{f_n\}$ converges uniformly on $E$
So, since $f+\hat{g}(x)^{n+1}+\sum_{i=0}^n\hat{f}_i(x)\hat{g}(x)^i=0$, I can evaluate the polynomial at $f$ and see that it must also be $0$ after evaluation because it's a ring homomorphism. I think that just finishes it off because the $f_i(x)$ were arbitrary
Ah, I have a problem. Just because $f_i(x)$ were arbitrary doesn't mean that $g(x)$ was arbitrary, so this doesn't necessarily account for every polynomial in $K[x]$, right?
@Rithaniel I think $F+xK[x]$ being integrally closed in $K[x]$ is equivalent to $F$ being algebraically closed in $K$ (in the sense of being equal to its own relative algebraic closure)
Here's how I would approach it. Suppose the subring is integrally closed. We need to show F equals its own relative algebraic closure in K. Take an element of K that is algebraic over F and note constants are polynomials too.
So, if we assume $F+xK[x]$ is integrally closed in $K[x]$, then let $k\in K$ is algebraic over $F$, then there exists $g(x)\in F[x]$ such that $g(k)=0$. However, since $F[x]\subseteq F+xK[x]$ (and this is where I lose it. $F+xK[x]$ being integrally closed means that I take any polynomial and can construct a larger polynomial out of it that ends up being the zero polynomial. That doesn't necessaily mean that I can conclude anything about the roots of these polynomials, right?) @Thorgott
This would be much easier if I knew what to google to get the right kind of input
If only there was a term for $F+xK[x]$
Like, I've written out "Suppose $F+xK[x]$ is integrally closed in $K[x]$ and let $k\in K$ be algebraic over $F$. Therefore there is a nonzero $g(x)\in F[x]$ such that $g(k)=0$. Since $F[x]\subseteq F+xK[x]$ then there exist $f_i(x)\in F+xK[x]$ such that $g(x)^n+\sum_{i=0}^{n-1}f_i(x)g(x)^i=0$" and I'm just staring at it because there is nowhere else to go, which I can see
Then I have to do the reverse implication, and then tackle four other problems: Let $F\subseteq K$ be fields. Show that the domain $F+xK[x]$ is completely integrally closed if and only if $F=K$ Show that $F+xK[x]$ is an HFD. Show that $F+xK[[x]]$ is an HFD. Show that the domains $F+xK[x]$ and $F+xK[[x]]$ are UFDs if and only if $F=K$.
Wait, I think I see it
Suppose $F+xK[x]$ is integrally closed in $K[x]$ and let $k\in K$ be algebraic over $F$. Therefore there is a nonzero $g(x)\in F[x]$ such that $g(k)=0$. Since $F[x]\subseteq F+xK[x]$ then the coefficients of $g(x)$ are polynomials in $F[x]$ and so $k$ is integral over $F[x]$ and must therefore be a constant polynomial in $F+xK[x]$, so $k\in F$. Therefore, $F$ is algebraically closed in $K$.
I thought what I was trying to prove is that $F$ is it's own algebraic closure. So if I have some $k\in K$ that is algebraic over $F$, I need to show that $k\in F$
Sure, yeah, I'll also post what I have on the work
Let $F\subseteq K$ be fields. Show that the domain $F+xK[x]$ is integrally closed if and only if $F$ is algebraically closed in $K$
$\implies$ Suppose $F+xK[x]$ is integrally closed in $K[x]$ and let $k\in K$ be algebraic over $F$. Therefore there is a nonzero $g(x)\in F[x]$ such that $g(k)=0$. We note that, since $F$ is a field, $g(x)$ is equivalent to a monic polynomial up to multiplication by a unit, so, WLOG, we assume $g(x)$ is a monic polynomial. Since $F[x]\subseteq F+xK[x]$ then the coefficients of $g(x)$ are polynomials in $F[x]$ and so $k$ is integral over $F[x]$ and must therefore be a constant polynomial in $F+xK[x]$, so $k\in F$. Therefore, $F$ is algebraically closed in $K$.
Well, we assume that $F+xK[x]$ is integrally closed, which means that if $k$ is integral over a subring of $F+xK[x]$, then it must be contained in $F+xK[x]$
Well, it's my understanding of it. When you take the integral closure of the ring, it has to remain inside the ring (if that ring is integrally closed). It stands to reason that the integral closure of a subring would also have to be inside the ring, as otherwise you'd arrive at a contradiction
$\implies$ Suppose $F+xK[x]$ is integrally closed (in $K(x)$ and therefore in $K[x]$), and let $k \in K$ be algebraic over $F$. We want to show $k \in F$. Now since $k \in K$ is algebraic over $F$, there is a nonzero $g \in F[x]$ such that $g(k) = 0$. By dividing by the leading coefficient, we can assume WLOG that $g$ is monic. Now interpreting $k$ as an element of $K[X]$ and $g$ as a monic polynomial with coefficients in $F+xK[x]$, we see that $k \in K[X]$ is integral over $F+xK[x]$,
which by assumption implies $k \in F+xK[x]$, so $k \in F$.
My mind wants to say "evaluate $g$ at the constant of $g$"
That gives us that the constant of $g$ is integral over $K[x]$, though, because the coefficients of $g$ are pulled from $K$
Got it (I think)
Suppose $F$ is algebraically closed in $K$ and let $g(x)\in K[x]$ be integral over $F+xK[x]$. We want to show that the constant term of $g(x)$ is in $F$. We know there exist $f_i\in F+xK[x]$ such that $g(x)^n+\sum_{i=0}^{n-1}f_i(x)g(x)^i=0$ and evaluating this polynomial at $x=0$ we get that $g^n+\sum_{i=0}^{n-1}f_ig^i=0$ where $g$ is the constant term of $g(x)$ and $f_i$ are the constant terms of $f_i(x)$. We note that this is a monic polynomial with coefficients in $F$ with $g$ as a root, and so we can immediately conclude that $g\in F$. Therefore $F+xK[x]$ is integrally closed over $K[x]$.
(Actually, I was apparently mistaken. It's just that the class group is most commonly used in algebraic number theory, so they usually define it on algebraic number fields.)
It seems to me 'mod' can be used in two senses. 1) as a binary operator 2) as a arithmetic system. Is this correct? In the equation above is it used in both senses?
Let $D_{\infty} = \langle a,t \mid a^2 = t^2 = 1 \rangle$ be the infinite dihedral group, and let $H = \langle at \rangle$. Then it isn't hard to verify that $D_{\infty}/H \cong \Bbb{Z}_2$. Why does this entail that $D_{\infty}$ is amenable?
Finite and abelian groups are easily shown to be amenable. So one might ask if all amenable groups can be built from finite and amenable groups using quotients, extensions, subgroups and direct limits. Turns out that the answer is no, groups that can be built this way are called elementary amenable
I've never seen a proper proof that there are amenable groups which are not elementary amenable, I think it's an hard result
Prescribe a map: $\Psi:\zeta_{\Bbb R^2} \to \Bbb T^2,$ which gives a transformation of $\zeta-$space in $\Bbb R^2,$ to the flat torus. Let $\zeta_{\Bbb R^2}$ consist of the following integral curves in which the source is $(1,0)$ and the sink is $(0,1)$: $ \phi_s(x)=e^{\frac{s}{\ln(x)}}. $ $x,\phi\in \Re(0,1)$ and $0\le \Re(s)<\infty.$ The integral curves $\phi_s(x)$ are solutions to the differential equation $X=\big(x\ln(x),-y\ln(y)\big).$ $\zeta_{\Bbb R^2}$ is then seen as a subregion of $X.$ How do I relate $X$ on $\zeta$ to some other vector field on $\Bbb T^2$? Is the push-forward viab…
I am reading "Intro to Manifolds" by Loring Tu, Ch. 14. Vector Fields. Tu states that in general it is very difficult to push-forward a vector field from one manifold to another. The concept of related vector fields is then discussed as a way to relate vector fields on different manifolds.
Hello everyone. I posted a question regarding some basic distribution theory but no one seems to answer. My course at university uses Kolmogorov and Fomin Introductory Real Analysis and specifically chapter 5 section 21 "Generalized functions". Is there someone available to perhaps help me with some basic questions regarding that part or perhaps point me some material so I can understand the material?
The functions are $\Bbb R\to\Bbb R$. I suspect being finite is a weird old way to say compactly supported, since the author obtains that they vanish outside of an interval from this property
If i want to construct a sequence of sets such that the first set lets call $A_0$ is the Naturals without the zero and the next set $A_1$ is the naturals without zero and one and the the next set $A_2$ is the naturals without 0 1 and 2 and so on and so forth, how would i write the discrebtion for it mathematically correct (the part with the removing elements like a premutation?) i cant figure it out! thanks!
@AlessandroCodenotti So when Kolmogorov writes: "Let K be the set of all finite (compact supported) functions $\phi$ from the real line into the real line such that they are continuously differentiable i.e. infinitely many time differentiable." It means in modern language that the functions are smooth and finite meaning compact support i.e. the support is $supp(f) = \{ x \in X : f(x) \neq 0 \}$ a compact subset of a topological space $X$ where in this case $X = \mathbb{R}$. And according to wiki
such functions are called bump functions. But how does compact support lead to the function vanishing outside some interval?
I know that compact in metric spaces is equivalent to closed and totally bounded. In particular for $\mathbb{R}^n$ this means closed and bounded.
@TedShifrin The flat torus is a torus with the metric inherited from its representation as the quotient, $\Bbb R^2 /L,$ where $L$ is a discrete subgroup of $R^2$ isomorphic to $Z^2$. This gives the quotient the structure of a Riemannian manifold. Right?
@TedShifrin sorry, can you write out what that means (in words), I just began learning this stuff
Is that "the pushforward of the vector field evaluated at a point $a$ equals the pushforward of the vector field evaluated at the point $p+\lambda$ for some constant $\lambda$?"
That's an involved question. Double points (nodes), triple points, cusps, tacnodes, etc. Being over $\Bbb C$ doesn't matter much except that you can draw pictures.
