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Jul
20
comment Miller-Rabin Primality Testing failure and a subgroup
For what it's worth, this result is stated as Exercise 3.17 in Crandall and Pomerance's book on primes.
Jul
18
comment Why are rational numbers required in cusps of congruence subgroups?
You are making a group act on the upper half plane. If you want to add infinity into the space then you sure better include the other points in its orbit under the group, and when ${\rm SL}_2(\mathbf Z)$ acts the orbit of $\infty$ aside from $\infty$ itself is in fact all the rational numbers.
Jul
18
comment Analogue of Fermat's primality test for polynomials and irreducibility
@0xbadf00d, can't you find examples yourself of different monic irreducibles of the same degree in some $\mathbf F_q[x]$ to check numerically that their product is a false positive?
Jul
18
comment Analogue of Fermat's primality test for polynomials and irreducibility
@0xbadf00d, read the proof of Korselt's criterion for Carmichael numbers and check it carries through to $\mathbf F_q[x]$ for any finite field $\mathbf F_q$: calling $f \in \mathbf F_q[x]$ "Carmichael" if $f$ is reducible and $a^{{\rm N}(f)-1} \equiv 1 \bmod f$ for all $a \in \mathbf F_q[x]$ relatively prime to $f$, where ${\rm N}(f) = q^{\deg f}$, a reducible $f$ is Carmichael iff it's squarefree and for any irreducible $\pi$ dividing $f$ we have $({\rm N}(\pi)-1)\mid ({\rm N}(f)-1)$. In particular, a product of two or more monic irreducibles of the same degree is Carmichael, as Jyrki wrote.
Jul
11
comment Finite (cardinality) modules over a PID
@MarianoSuárez-Alvarez, note the premise is that $R$ is a PID, so not every ring of integers in a number field is a candidate.
Jul
6
comment What does it even mean to say 'preserve structure'?
The property of being abelian ($xy = yx$ for all $x$ and $y$ in the group), the property of having an element of order 9, the property of having a subgroup of order 10, the property of having a normal subgroup of index 10, and the property of having a nontrivial commutator subgroup are all preserved by an isomorphism. You should sit down and try to check these yourself, e.g., if $f \colon G \rightarrow H$ is an isomorphism of groups then $G$ if abelian iff $H$ is abelian, $G$ has an element of order 9 iff $H$ does, and so on.
Jul
5
comment Is every field a Krull domain?
Let me ask you a related question: do you consider a field to be a UFD? A field has no irreducible elements (the set of irreducible elements is empty), so at best it is a degenerate example of a UFD and you would consider it to be a UFD just to have cleaner statements of theorems. In your concerns, an integral domain that is not a field does not become a Krull domain using an empty set of valuations since the valuations are defined on the fraction field, so the intersection in property 2 is the fraction field.
Jul
5
awarded  Good Answer
Jul
4
comment Symbol $\Gamma$ when talking about vector fields.
Yes, a section means each $m \in M$ is mapped to a point in its own tangent space $T_m(M)$, and that indeed is the idea behind a vector field.
Jul
4
comment Symbol $\Gamma$ when talking about vector fields.
@Hozer, no because a mere smooth map from $M$ to $TM$ does not have to map a point in $M$ to a point in its own tangent space. Take $M$ to be the unit circle $S^1$, whose tangent bundle is a cylinder. There are lots of ways to wrap a circle around a cylinder without respecting the natural map $T(S^1) \rightarrow S^1$.
Jul
4
comment Chance of adjacent lockers with the same combination
Since the number of actual Master Lock combinations is only 4000 (due to the mod 4 constraints $c\equiv a \bmod 4$ and $b \equiv a+2 \bmod 4$ on a Master Lock combination $(a,b,c)$ -- see the link in the comment by vadim123), the probability is even higher than what you wrote. Using 4000 in place of 60840 (you write 59280, which is an error by the OP since that is $40 \cdot 39 \cdot 38$) and taking a product for the complementary event of no shared combination among 1000 people, the probability of at least one shared combination among 1000 people is around $1−e^{−136}$.
Jul
4
comment Chance of adjacent lockers with the same combination
@vadim123, I agree that once you find a number in one of the positions (the last position in practice) there are only 100 possibilities for the overall combination, but the number you find in that position could be any of 0,1,...,39, so the overall number of combinations available in MasterLock space is $40\cdot 100 = 4000$.
Jul
4
comment Chance of adjacent lockers with the same combination
@vadim123, while the mod 4 conditions relating numbers in a Master Lock combination imply there are only 100 possible combinations with a particular (say) fist number, that first number can be any of 40 values, so the number of possible combinations is 4000, not 1000.
Jul
3
comment Generalized Sophomore's dream. Question about originality
Your English is generally good. Your main mistake is that you use commas too much. (Наверно Вы русский? Они люююбят запятые. :))
Jul
1
awarded  Nice Answer
Jun
30
comment The Frobenius Trace for an elliptic curve
See Ireland and Rosen's number theory book: Lemma 1 section 4 chapter 9 and Theorem 4 section 3 chapter 18.
Jun
27
comment Does an inseparable extension have a purely inseparable element?
Lipman wrote a short paper ("Balanced Field Extensions," Amer. Math. Monthly 73 (1966), 373-374) about the algebraic extensions that are separable extensions of inseparable extensions and at the end he gives a counterexample that is precisely this construction when $p = 2$ except he allows any field of characteristic $p$ in place of $\mathbf F_p$, which of course works in the above answer as well.
Jun
25
revised Help with proof that that affine plane curves in $\mathbb{C}^2$ are not compact
added 1 character in body
Jun
24
comment Alternate method to calculate an infinite string of numbers that's not $\pi$, and contains any string
What do you need this for? One answer mentions the Champernowne constant .12345..., which certainly works but has no more or less content than just listing all decimal strings of length 1, 2, 3,... in a specific order and can't really be "defined" in any way other than its decimal expansion, so it seems useless.
Jun
24
comment Example of two field extensions such that their tensor product is not a field
Your "map" is incorrect. It is just symbol-pushing and has no meaning. In fact your $K \otimes_k L$ is a field, namely $\mathbf Q(\sqrt{2},i)$. For an example try taking $L = K$.