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Aug 12 |
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isomorphism between specific generated field and specific quotient ring — gap in a proof of rings $$\hat{\varphi}:\frac{F[X]}{(h)}\to F[u]$$ : I repeat, $\hat{\varphi}$ is an isomorphism of rings. But from the first point I made and asked you to grant me for the moment, we can deduce that those are actually fields on both sides. Why? because the ideal $(h)$ is non zero, and the quotient ring $\frac{F[X]}{(h)}$ is isomorphic as rings to the subring $F[u]\subset K$ of the field $K$ which therefore is a domain. Hence $\frac{F[X]}{(h)}$ is a domain, and by the granted fact, it is a field. |
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Aug 12 |
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isomorphism between specific generated field and specific quotient ring — gap in a proof Thus, we have $h$ divides $\mu$, and $\mu$ divides $h$, they are both unitary polynomials with coefficients in a domain (actually the field $F$), so they are equal. This shows that $\ker(\varphi)=(h)=(\mu)$. Ok, by universal porperty, we get that there is an induced injective homomorphism of rings $\hat{\varphi}:\frac{F[X]}{(h)}=\frac{F[X]}{\ker(\varphi)}\to K$ whose image is the same as the image of $\varphi$, that is all of $F[u]$ by its definition and definition of $\varphi$. This yields, by restriction an injective, surjective ring homomorphism (I'll still call $\hat{\varphi}$) |
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Aug 12 |
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isomorphism between specific generated field and specific quotient ring — gap in a proof Since $\varphi(\mu)=\mu(u)=0$, we have $\mu\in \ker (\varphi)=(h)$ i.e. $h|\mu$. On the other hand, if $h(X)=Q(X)\mu(X)+R(X)$ is the euclidean division of $h$ by $\mu$, with $Q(X),R(X)\in F[X]$ and $\mathrm{deg} (R(X))<\mathrm{deg} (\mu(X))$ by definition of euclidean division, upon applying $\varphi$ to this equation you get $0=\varphi(h)=\varphi(\mu)\times\varphi(Q)+\varphi(R)=0\times\varphi(Q)+\varphi(R)=\varphi(R)$. Thus, $R$ annihiliates $u$ but has degree strictly smaller than $\mu$, the minimal polynomial of $u$, and thus $R=0$. This concludes that $h=Q\mu$ i.e. $\mu|h$. |
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Aug 12 |
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isomorphism between specific generated field and specific quotient ring — gap in a proof Ok, with this in mind, let's show that $(h)$ is prime, i.e., that the quotient ring $\frac{F[X]}{(h)}$ is a domain. Well, consider the homomorphism of $F$-algebras $\varphi:K[X]\to K,P(X)\mapsto P(u)$ (you call it $peval$). Since $u$ is algebraic over $F$, its kernel is a non-zero ideal of the PID $K[X]$, say $\ker(\varphi)=(h)$. $h$ is by definition a non zero polynomial, and without loss of generality, we can assume $h$ to be unitary, i.e. have leading coefficient equal to $1$. This $h$ is also the minimal polynomial of $u$ over $F$. Indeed, let $\mu(X)$ be $u$'s minimal polynomial. |
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Aug 12 |
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isomorphism between specific generated field and specific quotient ring — gap in a proof @boreal: dear boreal, I have taken time to answer your question, I think you should have the decency to point out what it is in my answer you don't understand. I argue that in order to show that the ideal $(h)$ is maximal, you only need to show that it is a prime ideal, because the ring $F[X]$ is a PID. This is a general theorem valid in all PIDs, a non-zero ideal in a PID is maximal iff it is prime. This can be seen easily in the special case where the PID is a polynomial ring with coefficients in a field. Grant me this for the moment being. |
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Aug 12 |
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isomorphism between specific generated field and specific quotient ring — gap in a proof @beroal have you even tried to conclude with what I wrote? The point is that you don't need to show $\ker(peval)$ is maximal, and I tell you exactly how to proceed! All you need to show is that the kernel is a prime ideal i.e. the quotient ring is a domain, which it is because it is isomorphic to a subring of a field... It's all written up in my answer. |
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Aug 11 |
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Differential forms on fuzzy manifolds Not an answer, and not addressing "fuzzy" manifolds, but differential forms can be defined on Banach manifolds and there's an analogue of the Frobenius integrality theorem (see Differential Forms by Cartan). However, I doubt this is what you're looking for. EDIT: This is most definitely not what you're looking for ^^ |
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Aug 11 |
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Groups of homeomorphisms of the real line @Theo Buehler but $f$ and $g$ don't commute: $g\circ f(1)=0$ while $f\circ g(1)=-1/2$. EDIT: I just realized you wrote non-abelian subgroups, my bad! |
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Aug 11 |
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Groups of homeomorphisms of the real line are there any non trivial relations you know of? |
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Aug 11 |
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What are topological manifolds for you? You need Hausdorff and second countability to get Paracompactness, and paracompactness and partitions of unity are very important for local to global constructions and in homotopy related questions : proving that homotopic maps induce isomorhpic pullbacks for instance. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval @Glenn Wheeler I have pointed out to you in a comment above that the OP gives the definition of differentiable he's working with. There is no $C^1$ hypothesis, only what the OP wrote down. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval @Glen Wheeler What? Let's define $f(x)=x^2\sin(1/x^2)$ for all non zero $x$ and $f(0)=0$ : $f$ is differentiable out side of $0$ for obvious reasons, and also at $0$, indeed for all non zero $x$, we have that $|x|<\epsilon$ implies $$\left|0-\frac{f(x)-f(0)}{x-0}\right|=|x\sin(1/x^2)|\leq|x|<\epsilon$$ which shows that $f$ is indeed differentiable at $0$ and $f'(0)=0$. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval @user9352 you previously wrote in a comment directed at me > "the idea is that a sequence of points say $t_n$ that approach $x$ from below are such that the derivative of $f$ at these points blows up to infinity". In the mean time, you edited your question and it reads > "show that $\lim_{t\to x^-}f'(t)\neq +\infty$ or $-\infty$ for all $x\in(a,b]$ These are different definitions : for the convenience of every one involved, please give the right definitions from the get go. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval @Glenn Wheeler I don't understand your point. The function I defined is differentiable in the usual sense, and indeed the sense of the question (user9352 gave his definition a few comments above). This is one of the most common counter example functions. What do you mean by $0$ can't be in $[a,b]$? |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval @Glenn Wheeler let me state what I forgot, set $f(0):=0$. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval If $\lim \dots =\pm\infty$ means $\lim \dots = +\infty$ or $\lim \dots =-\infty$, then this is not the hypothesis made by the OP. And in the general case, that is if you only make the hypothesis made by the OP : there is a sequence $y_n$ below $x$ such that $\lim_{\infty}f'(y_n)=+\infty$ (for instance), you will not have $\lim_{x^-}f'(c(y,x))=\lim_{x^-} f'(y)$. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval I didn't understand $1)$, which takes care of $f'(x)=\pm\infty$, but the other two points stand. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval Finally, $\lim_{y\to x^-} f'(c(y;x))\neq \lim_{y\to x^-} f'(y)$ simply because $c(y;x)$ has no incentive to visit all of $(x-\epsilon,x)$, what I mean is that there can be great gaps in the values taken by $c(y;x)$. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval There are several things wrong. First, the left limit is equal to $f'(x)$ since $f$ is differentialble. Thus it can't be equal to $\pm\infty$. Also $\lim f'(y)$ doesn't have to exist. |
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Aug 11 |
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derivative must be bounded on interior for a differentiable function on a closed interval I think your last line is incorrect, you need $x$ to move too in the left limit, otherwise this is false. |
