12,344 reputation
11332
bio website math.brown.edu/~lubinj
location Minnesota
age 77
visits member for 2 years, 9 months
seen 13 hours ago

Aged mathematician


Jun
10
comment Is C the only extension of R with degree 2?
Actually, $i$ is there: square your generator and subtract $2$.
Jun
9
comment Proving a trigonometric identity by infinite series
How far have you gotten?
Jun
8
comment Need Help Condensing This Logarithmic Expression
Don’t you suppose that OP meant $\log_5$?
Jun
5
comment History of the Coefficients of Elliptic Curves — Why $a_6$?
In confirmation of @zyx’s response, the sorting by lex order has nothing to do with it.
Jun
4
comment When to give up on a hard math problem?
The mind has a way of working when you’re not watching it. Sleep is an excellent way of letting things settle down. maybe to rearrange themselves in your mind.
Jun
4
comment Looking for a field isomorphic to $\Bbb{Q}$
Take $\mathbb Q$ and color all the elements blue. That’s as nontrivial as you can get.
Jun
3
comment Prove that if $p$ is prime then $\frac{X^p-1}{X-1}=X^{p-1}+\cdots+X+1$ is irreducbile in $\mathbb{Q}[X]$.
In a case like this you should always look at a few examples. The cases $p=2,3,5$ would show you immediately what the story is.
May
28
awarded  Enlightened
May
24
comment Transcendental Basis
Indeed, $\sqrt\pi$ isn't even in the (linear) span of OP’s set.
May
23
comment Complex no. $z$ satisfy the equation $|z-(1+i)| = 2$ , Then locus of $omega$ if $\omega = \frac{2}{z}$
Sorry for the delay. Your question is well answered by the response below. More generally, you can show for yourself the image of any circle under $z\mapsto (ax + b)/(cz + d )$ where $ ad -bc\ne0$ will be a circle or straight line. Same method.
May
16
comment $\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ Galois group
A much more advanced argument goes this way: $\mathbb Q(\zeta_9)$ is unramified at the prime $2$, but our total extension has cubic ramification there.
May
16
comment “And”/“or” in solutions of inequalities
Lang, so blameless in many other ways, was very sloppy.
May
16
comment Complex no. $z$ satisfy the equation $|z-(1+i)| = 2$ , Then locus of $omega$ if $\omega = \frac{2}{z}$
Did you sketch your circle? It’s closest to the origin where? And you know that the image of a circle under $z\mapsto2/z$ is a circle or straight line, don’t you?
May
15
comment Using two equations, creating a single equation with the same solutions as the original two?
wouldn’t $u=0$, $x=5$, $v=3$, $y=2$ show that your guess is wrong?
May
15
comment Does $\operatorname{arcsec}(x) = 1 /\arccos(x)$?
Remember that arcsec delivers an angle as its value. Ordinarily, it makes no sense to take the reciprocal of an angle.
May
15
answered What are primitive roots modulo n?
May
14
comment What does it mean to say that an element 'satisfies' a polynomial?
Purely as a matter of language, I would say that it’s ungrammatical or worse to say that an element satisfies a polynomial. Rather one should restrict the word “satisfy” to sentences with a free variable, like $f(x)=0$. That is, an element satisfies the sentence when the sentence becomes true when the variable is set equal to the named element.
May
14
comment Solving $\cos x=x$
“Why is there no exact solution” is asking the wrong question. The right question, when there happens to be an exact solution, is to ask why it does exist. That question will be answered by a proof. To ask why there are no manticores is the wrong question, but to ask why there are kangaroos is the right question.
May
14
comment K/F is finite separable extension show that N_K/F = N_E/F * N_K/E and Tr_K/F = Tr_E/F * Tr_K/E for any intermediate field K/E/F.
This is not multiplication of values, but composition of functions. The Norm $N^K_E$ takes elements of $K$ and gives values in $E$; then $N^E_F$ takes an element of $E$ and gives you an element of $F$.
May
13
answered Proving the p-adic numbers $\mathbb{Q}_p$ form a field