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Apr
28
comment For any four distinct points $P_{1},P_{2},P_{3},P_{4}\in P^{1}(\mathbb{R})$ we have $(P_{1}P_{3};P_{2}P_{4})=1-(P_{1}P_{2};P_{3}P_{4})$.
Since the cross-ratio is given by a formula, couldn't you prove it directly?
Apr
28
comment Is the size of the conjugacy class of a given element in a compact Lie group always finite?
For a geometric example, the conjugacy class of a nontrivial rotation $R$ in ${\rm SO}(3)$ by an angle of $\theta$ is the set of all rotations around axes by an angle of $\theta$. (Indeed, if $R$ rotates around the oriented axis $\ell$ by $\theta$, then $SRS^{-1}$ rotates around the oriented axis $S\ell$ by angle $\theta$, and ${\rm SO}(3)$ acts transitively on oriented axes.)
Apr
26
comment Lattice of a subfield
You should read some lecture notes on finite fields. It is still not clear to me you know what $GF(p^n)$ means, or what a lattice of fields is.
Apr
26
comment Lattice of a subfield
Where did you see $GF(24)$? Are you sure it wasn't $GF(2^4)$ or maybe $GF(2,4)$? In any case, I don't see why you're asking for the answer to the lattice question when you should be first asking what $GF(p^n)$ means. You need to build up basic knowledge first before you can tackle these sorts of exercises.
Apr
26
comment Lattice of a subfield
The only $a$ for which $GF(a)$ makes sense are those $a$ which are powers of primes. So your claim that you've seen these exercises but never seen something like this is a contradiction. Actually GF stands for Galois field, aka finite field. Not quite the same thing as prime field.
Apr
26
comment Lattice of a subfield
Yes $p$ is a prime. Do you know $GF(p^3)$ means?
Apr
24
comment The ideals $\langle y-x-1\rangle$ and $\langle x-2,y-3\rangle$ in $\mathbb C[x,y]$ are prime
Yes to your questions at the very end. They are kernels of evaluation maps.
Apr
23
comment Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
I see no requirement for separable polynomials in the definition of splitting field. If you're asking why we have that requirement to call the splitting field of an irreducible polynomial a Galois extension, it's because we want "enough" symmetries but there aren't "enough" symmetries if the polynomial is inseparable. Most advanced books or lecture notes on number theory, or field theory or even Galois theory should cover purely inseparable extensions. You can peruse Pete L. Clark's online notes, for example.
Apr
23
comment Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
Sorry, I was thinking irreducible on top of inseparable. In any case, you're familiar with the non-Galois inseparable extension $\Bbb F_q(t)/\Bbb F_q(t^p)$ right?
Apr
23
comment Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
In any case, to address the question in the title, if $K/F$ is the splitting field of an inseparable polynomial over $F$, then $K/F$ is automatically not Galois.
Apr
23
comment Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
Any product of linear polynomials over $F$ will split in $K$, no matter what $K/F$ is. In particular, we will always be able to say $x^2-x$ is separable and splits in a field $K$.
Apr
23
comment What does x equivalent to 2 mod 15 mean?
Computer scientists and programmers treat mod as a binary operation, taking in two inputs and producing one output. Mathematicians treat modulo as a family of binary relations between two numbers, one such relation for each possible modulus. The relation $a\equiv b$ mod $n$ means that $a$ and $b$ differ by a multiple of $n$, or in other words that the difference $a-b$ is a multiple of $n$, or equivalently that $a$ and $b$ have the same remainder upon division by $n$.
Apr
20
awarded  Enlightened
Apr
20
awarded  Nice Answer
Apr
16
comment Show that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
Can $\sqrt[3]{5}^2$ be expressed like that?
Apr
16
comment Where can I read about this 'rule'?
If $c$ is a $1\times 1$ matrix, then $c^T=c$. Set $c=a^TXb$. Then $(a^TXb)^T=b^T X^T (a^T)^T=b^TX^Ta$.
Apr
14
comment Why do division algebras always have a number of dimensions which is a power of $2$?
The sedenions are not a division algebra. Also, it may help to distinguish division algebras from composition algebras (normed division algebras). Division algebras (over $\Bbb R$) are more plentiful (although still restricted to the same dimensions 1,2,4,8) and considerably harder to prove things about than composition algebras.
Apr
14
comment What is a simple means of proving that 3 vectors belonging to $\Bbb{R}^2$ are linearly dependent?
Seems overly complicated. Why not just mention dimensions and bases/geometry?
Apr
14
comment Proving that a Galois group is cyclic
The extension $K(t)/K(t^n)$ is just $F(\alpha)/F$ where $F=K(t^n)$ and $\alpha$ is a root of $X^n-t^n$.
Mar
15
awarded  Enlightened