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May
2
comment What did Alan Turing mean when he said he didn't fully understand dy/dx?
It's funny to me that Turing thinks it is too subtle to treat polynomials with coefficients in $K$ as formal expressions and prefers to regard them as functions $K \rightarrow K$, considering the importance of finite fields in computer science. They are exactly the fields where this change in point of view really wrecks the algebraic structure of polynomials.
May
1
comment Physical significance and graphical point of view of second derivative of a function $f''(x)$ .
There is no such thing as a "double derivative." It's like asking why in a sequence $a_1, a_2,\dots$ we don't call $a_2$ the "double term." It's just not the term that is used.
May
1
comment Physical significance and graphical point of view of second derivative of a function $f''(x)$ .
Your question is about the second derivative (it is not called the "double derivative") but the subject line ends with $f'''$. Fix that.
Apr
30
comment Why did mathematicians introduce the concept of uniform continuity?
If you want to see the difference between uniform convergence and pointwise convergence, look into the Gibbs phenomenon.
Apr
30
comment Automorphism of $\mathbb{Q}({\zeta_n})/\mathbb{Q}$
Your question does not make much sense, since ${\rm Gal}(\mathbf Q(\zeta_n)/\mathbf Q)$ and ${\rm Aut}(\mathbf Q(\zeta_n)/\mathbf Q)$ are exactly the same thing. So it seems to me that you are asking "is this true if it is true?"
Apr
30
comment Maximal ideals of polynomial ring $\mathbb Z_n[x]$
Take the quotient by that and see if you get a field. That would tell you if it is maximal. For comparison, is $p\mathbf Z$ a maximal ideal in $\mathbf Z[x]$?
Apr
30
comment Maximal ideals of polynomial ring $\mathbb Z_n[x]$
If you understand things well enough, then certainly. (I guess you don't if you had to ask.) You just need to figure out which maximal ideals in that ring contain your $n$. After all, the ring you write is isomorphic to $\mathbf Z[x]/(n)$, so you can use the description of maximal ideals in a quotient ring $R/I$ in terms of maximal ideals in $R$. But there can be other solutions that don't rely on knowing maximal ideals of $\mathbf Z[x]$.
Apr
30
comment Maximal ideals of polynomial ring $\mathbb Z_n[x]$
Do you know what the maximal ideals of $\mathbf Z[x]$ look like?
Apr
29
comment Can all “standard” properties of the tensor product be proven from the universal property?
Even if the answer is yes, don't forget that "existence" is a pretty important property too.
Apr
29
comment Finding roots in finite fields.
@Lubin, I agree and when I realized there is such a method later I forgot what I had first written above.
Apr
27
awarded  Necromancer
Apr
27
comment Is the discriminant of a polynomial surjective onto $\mathbb Z$?
It is a theorem of Stickelberger that any discriminant is $0$ or $1 \bmod 4$. And not all values satisfying one of those congruence conditions are necessarily possible, e.g., no polynomial in $\mathbf Z[x]$ of degree at least $2$ has discriminant $1$. A related question: math.stackexchange.com/questions/820990/….
Apr
27
comment Proving that $D_{12}\cong S_3 \times C_2$
There is a more general result in this direction: if $n \geq 6$ is twice an odd number then $D_{2n} \cong D_n \times C_2$. Your question is about $n = 6$. Note $D_6 \cong S_3$ in the notation you are using.
Apr
27
comment Showing a certain cyclotomic polynomial must split
The original proof by Artin and Schreier uses cyclotomic polynomials, but that can be avoided. See math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf.
Apr
26
comment Let $F,K$ be two fields $F \subset K$ and suppose $f(x),g(x) \in F[x]$ are relatively prime in $F[x].$ Prove they are relatively prime in $K[x].$
The more interesting aspect is that the converse is true: if $f$ and $g$ in $F[x]$ are relatively prime in $K[x]$, where $F \subset K$, then they are relatively prime in $F[x]$.
Apr
26
comment Reducing mod $n$ in $\mathcal O_{\mathbb Q (\zeta_n)}$
A fourth-year project? Just read a book on algebraic number theory (e.g, Samuel's "Algebraic Number Theory" and learn more about it. There's no need to make it a formal "project."
Apr
26
comment Reducing mod $n$ in $\mathcal O_{\mathbb Q (\zeta_n)}$
More simply, in a quadratic field $K = \mathbf Q(\sqrt{d})$ like your two examples, the way an unramified odd prime $p$ decomposes in $\mathcal O_K$ is related to how $x^2 - d \bmod p$ decomposes, which is related to the Legendre symbol $(\frac{d}{p})$. The quadratic reciprocity law implies that $(\frac{d}{p})$ for fixed $d$ is determined by $p \bmod 4d$; when $d \equiv 1 \bmod 4$ the Legendre symbol $(\frac{d}{p})$ is even determined by $p \bmod d$. Taking $d = -1$ and $d = -3$ recovers your observation about the role of $p \bmod 4$ (same as mod $-4$) and $p \bmod 3$ (same as mod $-3$).
Apr
26
comment Reducing mod $n$ in $\mathcal O_{\mathbb Q (\zeta_n)}$
You're right in to guess that there's something about $p \bmod n$ (what you call $n$-ness) that tells you how the ideal $(p)$ factors in $\mathbf Z[\zeta_n]$. It's because $\mathbf Q(\zeta_n)/\mathbf Q$ is a Galois extension with Galois group isomorphic to $(\mathbf Z/(n))^\times$, and for a Galois extension $K/\mathbf Q$ there is a close relation between how an unramified prime $p$ factors in $K$ and a certain conjugacy class in ${\rm Gal}(K/\mathbf Q)$ related to $p$, called its Frobenius conjugacy class. If the Galois group is abelian, the conjugacy class becomes an element.
Apr
26
comment Reducing mod $n$ in $\mathcal O_{\mathbb Q (\zeta_n)}$
In $\mathbf Z[\zeta_8]$, for instance, if $p \equiv 1 \bmod 8$ then the ideal $(p)$ splits into a product of four prime ideals and for all other odd prime numbers $p$ the ideal $(p)$ is a product of two prime ideals (each with norm $p^2$). No prime number stays prime in $\mathbf Z[\zeta_8]$.
Apr
26
comment Reducing mod $n$ in $\mathcal O_{\mathbb Q (\zeta_n)}$
The rings for $n = 3$ and $n = 4$ are quadratic, so the only choice a prime has is to stay prime or split (or ramify, but that happens once, for $p = 2$ in $\mathbf Z[i]$ and $p = 3$ in $\mathbf Z[\omega]$). For the ring of integers $\mathbf Z[\zeta_n]$ of a general cyclotomic field, the splitting behavior of a prime number $p$ that doesn't divide $n$ is determined by the subgroup $\langle p \bmod n\rangle$ of $(\mathbf Z/(n))^\times$. In particular, $p$ is prime in $\mathbf Z[\zeta_n]$ iff $p \bmod n$ generates $(\mathbf Z/(n))^\times$. Often there's no such $p$, e.g., $n = 8$ or $15$.