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Jun
13
comment $S$ subring of $R$. Is a projective objects in $R$-$\bmod$ still projective in $S$-$\bmod$?
As another comment, $\mathbb{Q}$ is a flat $\mathbb{Z}$-module, which is maybe a helpful exercise
Jun
13
comment $S$ subring of $R$. Is a projective objects in $R$-$\bmod$ still projective in $S$-$\bmod$?
@Biagio: There are a couple of ways to see that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module. See for example math.stackexchange.com/questions/506256/….
Jun
7
comment show set is prime ideal
In addition to mich95's method, you can do this 'by hand'. If $(a,b)(c,d) \in I$ then $bd = 0$ so $b=0$ or $d=0$ because $\mathbb{Z}$ is an integral domain.
Jun
7
answered What does it mean if the derivative of a function is a constant?
Jun
7
comment Prove that if the set of ideals is $\{\{0\}, R \}$, then $R$ is a field.
Yes I might be circular here. I have seen some authors use this as the definition of maximal, though.
Jun
7
comment If $f$ is a group homomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Q}-\{0\},.)$ such that $f(2)=\frac{1}{3}$, then find $f(-8)$.
@Ritu: Yes. If $f$ is a homomorphism then $f(x^{-1}) = f(x)^{-1}$ (I have used multiplicative notation so that we don't worry about commutativity). The proof is the same as the one Thomas gave, only with the different notation.
Jun
7
comment If $f$ is a group homomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Q}-\{0\},.)$ such that $f(2)=\frac{1}{3}$, then find $f(-8)$.
To strengthen mich95's hint, if $f$ is a homomorphism of groups, what is $f(-x)$?
Jun
7
answered Prove that if the set of ideals is $\{\{0\}, R \}$, then $R$ is a field.
Jun
4
comment Finding $\lim_{n\rightarrow \infty }\left(1-\frac{1}{n^2}\right)^n$
What have you tried? Hint: consider using logarithms.
May
2
comment How are singularities of complex functions classified?
No, it's easier with rational functions like this one. You have a pole at $z_0$ when $f(z)(z-z_0)^n$ is analytic and nonzero for some positive integer $n$.
May
2
comment How are singularities of complex functions classified?
Remember that there are three possibilities for singularities: removable, pole, essential. You seem to have ruled out removable. Do you know what a pole is?
May
2
awarded  Autobiographer
Apr
18
comment Integrating a vector field over curve in R^2 with differential forms
Thanks. I guess that's pretty straightforward.
Apr
18
asked Integrating a vector field over curve in R^2 with differential forms
Oct
24
comment Tips for finding the Galois Group of a given polynomial
I think it's helpful to first consider the size of the Galois group, then ask which group of that size it could be. For example, if $|G|=6$, then it suffices to figure out whether $G$ is abelian
Oct
24
answered A question on Newton's “theorem about ovals”
Oct
24
comment A question on Newton's “theorem about ovals”
Or I should ask -- at the end of your answer, you say "For sufficiently smooth curves, if the area cut off is algebraic at all, it has to be algebraic over the whole domain of the angle." Why must it be algebraic over the whole domain of the angle?
Oct
24
comment A question on Newton's “theorem about ovals”
Isn't that precisely the definition of "$f$ is an algebraic function"? Further, I wouldn't claim $f$ and $g$ measure the same thing, merely that they agree for $\theta\in[2\pi ]$. For example, if we take the unit circle and a ray from the origin making angle $\theta$ with the poisitive $x-$axis, I would say that the area of the sector is $f(\theta) = \theta$ for $\theta \in [0, 2\pi)$ and is continued to the real line via $f(\theta + 2\pi) = f(\theta).$
Oct
23
comment A question on Newton's “theorem about ovals”
Ok, then I agree. But I still have my original question. The theorem states, as I understand it, that there is no polynomial $P$ with $P(f,r,s,t)=0$, where $f(r,s,t)$ is the area of the curve cut off by the line $rx+sy=t.$ The proof seems to provide a function $g(r,s,t)$ which has that property, but $g\neq f$
Oct
23
comment A question on Newton's “theorem about ovals”
How can a function be both periodic and monotone (and not be constant)?