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1d
comment Field Theory Problem in Beachy's Abstract Algebra involving field extensions and transcendental elements.
You should write $K(u)$, not $K[u]$. The first notation is a field, the second is the polynomials in $u$ with coefficients in $K$.
2d
comment Where did the German term “Spur” of a matrix come from?
This issue came up on the site earlier. See the comments to the answer by Bernard at math.stackexchange.com/questions/1086207/…
2d
comment Is the semidirect product of normal complementary subgroups a direct product.
It is a basic exercise to check in a general semidirect product $K \rtimes_{\varphi} N$ that the subgroup $N$ (identified with pairs $(e,n)$ for $n \in N$) is a normal subgroup if and only if $\varphi$ is trivial: $\varphi_n : K \rightarrow K$ is the identity on $K$ for all $n \in N$.
2d
comment Is every reduced $k$-algebra all of whose residue fields are $k$ finitely generated?
If $k$ is a finite field of order $q$ then $x^q = x$ for all $x$ in a direct product of copies of $k$, so any quotient integral domain must be $k$. Therefore in that special case your guess about a problem coming from nonprincipal ultrafilters does not occur.
May
20
comment Show that the ring $R$ of entire functions does not form a Unique Factorization Domain
It's a basic theorem in complex analysis that for any open connected subset $\Omega$ in $\mathbf C$, the holomomorphic functions on $\Omega$ are an integral domain. In particular, this is true when $\Omega = \mathbf C$. This should be discussed in lots of complex analysis books.
May
19
awarded  Nice Answer
May
16
comment What are the elements with the minimal absolute values in cyclotomic extensions of the integer ring?
The number $|a|$ is not a norm. Algebraic integers have norms in $\mathbf Z$ so the minimal nonzero norm in absolute value is $1$, which is the norm of $1$.
May
16
comment Groups of order 56 with a normal Sylow 7-subgroup
Isomorphic groups can be described by generators with different relations! There is no contradiction anywhere.
May
13
comment $S_4$ is not supersolvable? Why am I wrong?
@RolfHoyer, that does not prove $S_4$ is not supersolvable, but only that this particular series of subgroups does not fit the condition for showing $S_4$ is supersolvable. Maybe another series of subgroups would fit the definition. A reason $S_4$ is not supersolvable is that its only normal subgroups are $\{(1)\}$, $K$, $A_4$, and $S_4$. In particular, since none of these groups (besides $\{(1)\}$) are cyclic, there is no normal series for $S_4$ whose successive quotient groups are all cyclic.
May
11
comment Galois extension over power series fields
That $L((X))$, which is $X$-adically complete, is not necessarily algebraic over $K((X))$ when $L/K$ is algebraic is similar to the proof that the algebraic closure of $\mathbf Q_p$ is not complete:an infinite series $\sum a_np^n$ where $a_n$ is a root of unity of order $p^n-1$ lies in the completion of the algebraic closure of $\mathbf Q_p$ but is not in the algebraic closure itself.
May
10
comment Expressing $\frac {x^n}{(1-x)^n}$ as a generating function
Please reread what you wrote and fix it. The $n$ on the left can't be the same as the $n$ in $x^n$ on the right. Otherwise the right side is a series where everything has the same term $x^n$.
May
10
comment Action of $SL_2(p)$ on intgers mod p by Möbius transformations
I'd say in $\mathbf F_p$ that it literally is division too. In any field $a/b = ab^{-1}$ (when $b \not= 0$). But I understand what you mean.
May
10
comment Action of $SL_2(p)$ on intgers mod p by Möbius transformations
The answer will all of the time be in this set. Please give even one counterexample.
May
9
comment Ordinary Differential Equations with undetermined coefficient
Is the right side supposed to be $7t^5$? Please type exponents and subscripts so they all appear in the right place.
May
8
comment The splitting field of $x^7 - 2$ and its Galois group
The degree of a nontrivial $7$th root of unity over $\mathbf Q$ is not $7$, but $6$. Since $[\mathbf Q(\sqrt[7]{2}):\mathbf Q] = 7$, which is relatively prime to $6$, the degree of a nontrivial $7$th root of unity over $\mathbf Q(\sqrt[7]{2})$ is also $6$. The answer sheet was not wrong. You are also making a bad deduction that if $\alpha$ and $\beta$ have the same degree $m$ over $\mathbf Q$ then $\mathbf Q(\alpha,\beta)$ has degree $m^2$ over $\mathbf Q$. That is just false in general (but it's also not pertinent to your situation if it is done correctly).
May
8
comment A paper on the relation between Ernst Kummer's work on factorization and structural linguistics (Zellig Harris)
There is no question here. Promoting your own work as a post is not really the purpose of this site.
May
8
comment Question About Morera's Theorem
It is not the minimum. It suffices for the integral to vanish on triangles. See the end of the Wikipedia page on Morera's theorem.
May
8
revised Question About Morera's Theorem
added 10 characters in body; edited title
May
8
comment Why does $N(\mathfrak p)$ belong to $\mathfrak p$?
Why in a finite field $F$ of order $q$ does $q = 0$ in $F$? Think about the characteristic.
May
8
comment Is $x^4+nx+1$ irreducible?
If $n$ is odd the polynomial is irreducible mod 2. If $n$ is a multiple of $4$ then $f(x+1)$ is Eisenstein at 2. This still leaves $n$ that are even but not a multiple of $4$.