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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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Aug
11
revised The quotient of the ring of integers by an ideal always seems to have the same structure
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Aug
11
answered The quotient of the ring of integers by an ideal always seems to have the same structure
Aug
9
comment How do you quickly find the eigenvalues of this matrix?
The numbers in your matrix are kind of large for hand calculation, so why do you think it is a good representative example of what could be on the exam? Is it taken from a real exam review or is it from elsewhere? Have you heard of the characteristic polynomial of a square matrix?
Aug
8
comment Discriminant of a trinomial
What you defined is not the usual definition of the discriminant: in the quadratic case your definition is $4c-b^2$, so there is a sign error. In general your definition of the discriminant is missing the factor $(-1)^{n(n-1)/2}$. I mention this in case anyone comes across this page and tries to use the boxed formula in the question as a discriminant formula for $x^n - bx + c$ without checking it themselves. They will likely be making a sign error (depending on $n \bmod 4$).
Aug
4
awarded  Yearling
Aug
3
comment On isomorphisms of tensors of certain type
When you write $\sigma \circ f \circ \sigma^{-1}$, your two $\sigma$ maps are not really the same (one on $V_K$, one on $W_K$). Why not think of your $\sigma(f)$ as a composite of three isomorphisms between appropriate spaces?
Aug
3
comment On the conjugates of $1-e^{\frac{2\pi i}m}$
The conjugates of a number over $\mathbf Q$ are the roots of that number's minimal polyomial over $\mathbf Q$. So if $f(x)$ is the minimal polynomial of a number $\omega$, what is the minimal polynomial of $1-\omega$? It is $(-1)^nf(1-x)$, where $n = \deg f$. Without using explicit minimal polynomials, the degree of a number over $\mathbf Q$ is the dimension of the field it generates over $\mathbf Q$. Since $\mathbf Q(\omega) = \mathbf Q(1-\omega)$, $\omega$ and $1-\omega$ have the same degree over $\mathbf Q$, as do $\omega$ and $a\omega + b$ for any rational $a$ and $b$ with $a \not= 0$.
Aug
2
comment dimension of $\mathbb{C}$ over $\mathbb{Q}$
It's an extension of infinite degree, of course. A finite-degree extension of the rationals would be countable.
Aug
2
comment Why does the order of summation of the terms of an infinite series influence its value?
Absolute convergence not only justifies that series rearrangements don't affect the sum, but this condition exactly captures that property: any series of real numbers that is not absolutely convergent can be rearranged to converge to anything at all (i.e., the order of the terms in the series generally matters). This is what makes Fourier series generally so subtle, since such series can be convergent but not absolutely convergent. That absolutely convergent series are those that can be rearranged without changing the sum is one reason the absolute convergence hypothesis appears a lot.
Aug
2
comment Why does the order of summation of the terms of an infinite series influence its value?
The higher-dimensional version of this phenomenon is addressed at mathoverflow.net/questions/29333/…
Jul
24
answered Solutions to the Mordell Equation modulo $p$
Jul
22
comment Number of monic irreducible polynomials over a finite field
This sounds completely backwards. The typical way to derive that formula for $\nu_n$ is to start from the formula $q^n = \sum_{k|n} k\nu_k$. If you're trying to derive this from the formula for $\nu_n$, then how have you obtained the formula for $\nu_n$ in the first place?
Jul
22
revised Finding non-negative integers $m$ such that $(1 + \sqrt{-2})^m$ has real part $\pm 1$.
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Jul
22
revised Finding non-negative integers $m$ such that $(1 + \sqrt{-2})^m$ has real part $\pm 1$.
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Jul
22
revised Finding non-negative integers $m$ such that $(1 + \sqrt{-2})^m$ has real part $\pm 1$.
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Jul
22
revised Finding non-negative integers $m$ such that $(1 + \sqrt{-2})^m$ has real part $\pm 1$.
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Jul
21
comment Finding non-negative integers $m$ such that $(1 + \sqrt{-2})^m$ has real part $\pm 1$.
Setting $b_m = a_m - 1$ and $c_m = a_m + 1$, the recursions are $b_m = 3b_{m-1} - 5b_{m-2} + 3b_{m-3}$ and $c_m = 3c_{m-1} - 5c_{m-2} + 3c_{m-3}$ (the same recursions) with different initial conditions: $b_0 = 0$, $b_1 = 0$, and $b_2 = -2$, while $c_0 = 2$, $c_1 = 2$, and $c_2 = 0$. The $2$-adic approach in my answer below does not need these recursions, as it starts from the exponential formula for the sequence $a_m$; I am mentioning this recursion issue only because your setup (seeking values $\pm 1$) is not how you usually see Skolem-Mahler-Lech stated in the literature (value 0).