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5h
comment Unfamiliar notation. (actuarial science)
You are a math instructor and you do not know what min means? Anyway, it would help if you gave more context, e.g., you write $d$ but do not say what it is.
6h
comment Characterization of continuity in terms of preimages of open sets
To me it is obvious that you reading Spivak's Cakculus on Manifolds, although you do not explicitly say so. What exactly is your background in real analysis? Do you know what a metric space is?
6h
comment Characterization of continuity in terms of preimages of open sets
The choice of $V$ depends on $U$ (and $f$), so you are just not picking $V$ well.
Aug
27
comment Proving that $A$ is diagonalizable
If $k < n$ then $X^k-1$ certainly won't have $n$ roots, but also it is not necessary to have all distinct eigenvalues to be diagonalizable. Just consider the identity matrix!
Aug
24
comment Does every group have a 'cyclization'?
Bryan, actually the direct sum $\bigoplus_{k \geq 1} \mathbf Z/p^k\mathbf Z$ is also an example. It has no nonzero homomorphism to $\mathbf Z$ since its elements are all of finite order, and it has no finite cyclization since it has arbitrarily large finite quotients.
Aug
24
comment Does every group have a 'cyclization'?
@AlexYoucis: swap the direction of the maps. :)
Aug
24
comment Does every group have a 'cyclization'?
@AlexYoucis: I was trying to give an argument eliminating each prime other than $p$ separately, while you use them all together (you appeal to there being infinitely many primes). Your argument also can be made to work prime-by-prime by looking at $q^m$-divisibility for fixed $q \not= p$ and arbitrarily large $m$.
Aug
24
comment Does every group have a 'cyclization'?
@AlexYoucis: Yes, your argument is simpler. I was trying to work out an argument myself before I read your comment as closely as I should have.
Aug
24
revised Does every group have a 'cyclization'?
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Aug
24
revised Does every group have a 'cyclization'?
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Aug
24
revised Does every group have a 'cyclization'?
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Aug
24
answered Does every group have a 'cyclization'?
Aug
23
revised What is $\hat{\mathbb{Z}}$?
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Aug
23
revised Is 2D FFT separable?
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Aug
20
comment Prove that a UFD $R$ is a PID if and only if every nonzero prime ideal in $R$ is maximal
For a correct proof that if all prime ideals are principal (Lemma 2 is true) then all ideals are principal, see math.stackexchange.com/questions/168082/…
Aug
19
comment Irrational number “test”?
@user1729: Surely you meant it's undecidable in general whether a number is rational or not, not whether it's real or not. Otherwise I'd wonder what you could have in mind by the phrase "given a number".
Aug
14
comment The quotient of the ring of integers by an ideal always seems to have the same structure
When is the size of $\mathcal O_K/\mathfrak p^r$ equal to $p^r$, where $p$ is the prime number that $\mathfrak p$ lies over? For any nonzero prime ideal $\mathfrak p$ and any $r \geq 1$, the general formula for the order of that ring is $p^{fr}$, where $f = f(\mathfrak p|p)$ is the residue field degree of $\mathfrak p$ over $p$. Therefore the order of that ring is $p^r$ if and only if $f = 1$, meaning that $\mathcal O_K/\mathfrak p$ has order $p$ (rather than order some higher power of $p$).
Aug
14
comment The quotient of the ring of integers by an ideal always seems to have the same structure
Any finite ring is a unique product of finite local rings. See the discussion of Artin rings in Atiyah and Macdonald's book on commutative algebra; any finite ring is an Artin ring, so the decomposition described there for Artin rings as a direct product of Artin local rings can be applied to the case of finite commutative rings.
Aug
14
comment The quotient of the ring of integers by an ideal always seems to have the same structure
Let's just forget about ramification and see what happens in general. Just as $\mathbf Z/(m)$ breaks up into the direct product of the local rings $\mathbf Z/(p^{e_i})$, where $p_1^{e_1}\cdots p_k^{e_k}$ is the prime factorization of $m$, for any number field $K$ and nonzero ideal $\mathfrak a$ in $\mathcal O_K$ the finite ring $\mathcal O_K/\mathfrak a$ breaks up into the direct product of local rings $\mathcal O_K/\mathfrak p_i^{e_i}$, where $\mathfrak p_1^{e_1}\cdots \mathfrak p_k^{e_k}$ is the prime ideal factorization of $\mathfrak a$.
Aug
12
revised The quotient of the ring of integers by an ideal always seems to have the same structure
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