Reputation
10,186
Next privilege 15,000 Rep.
Protect questions
Badges
23 51
Newest
 Necromancer
Impact
~207k people reached

Apr
19
comment Can one define a functional on a Hilbert space based on its action on a Hilbert basis?
You need a continuity condition as well. Simply assigning any random values you want to the Hilbert space basis is not going to extend to a continuous linear map. Do you believe that two (continuous) linear functionals that are equal on a Hilbert space basis are equal everywhere?
Apr
19
comment Research Paper on Math
I realized that, but since you ultimately need to be writing a paper I wanted to point out that describing an equation as American or non-American is simply bad writing. Always be attentive to how you express your ideas.
Apr
19
comment differentiability, complex analysis
Try working with some concrete examples (polynomials, rational functions) to see what a likely formula for the derivative of $f^*(z)$ should be. I personally think this is a really good problem.
Apr
19
comment If $R$ is a ring with unit element $1$ and $\varphi$ is a homomorphism of $R$ onto $R'$, prove that $\varphi(1)$ is the unit element of $R'$.
Here is a concrete example: let $R = R' = {\mathbf Z}/(6)$. Let $f(x) = 3x$. Then $f(x)$ is additive and multiplicative (since $3^2 \equiv 3 \bmod 6$), but $f(1) \not= 1$. Note $f$ is not surjective. You should be made aware that the standard convention of most mathematicians who use algebra in their work (esp. anyone who uses commutative algebra) is that by definition a ring homomorphism must satisfy $f(1) = 1$. That is, most mathematicians who use algebra would not call a general additive/multiplicative map between rings a ring homomorphism unless it satisfies $f(1) = 1$.
Apr
19
comment Research Paper on Math
Calling equations "non-American" makes me cringe. An equation does not have a nationality. (And you might want to read up on the journal Deutsche Mathemtatik.) Please watch how you express your ideas when you get around to writing the paper itself.
Apr
19
comment Composition of Polynomials and Galois Theory
A Google search on "composition polynomials galois" returned the article ccms.or.kr/data/pdfpaper/jcms22_3/22_3_497.pdf as one of the top hits (did you try an internet search before asking your question?) and the paragraph after Lemma 3.2 will be of interest to you.
Apr
19
comment Is Euclidean geometry really a “dead” subject? If so, why?
Your general premise is wrong. My university has a Euclidean and non-Euclidean geometry course (taken mostly by future high school math teachers). You say Harvard doesn't seem to have such a course, but did you see their Math 130? Though that is not a course aimed at the superior math majors. It is not exciting math compared with all the other topics that a math major could study (differential geometry, complex analysis, ...)
Apr
19
comment Is Euclidean geometry really a “dead” subject? If so, why?
I think the question is still pretty rant-y. Euclidean geometry is not isolated. It extends to higher dimensions and in different directions (contrasts with non-Euclidean geometry). Math is a verrrrry big subject, and just as nobody learns an algorithm for computing cube roots by hand anymore (I barely got a glimpse of that when I was in high school) other material taught for a long time will have to give way so that more contemporary topics can be discussed. Latin and Greek used to be a standard part of everyone's higher education, but now....
Apr
19
comment Provide an example to show that $S$ may not necessarily be a unique factorisation domain when $R$ is a unique factorisation domain.
Did you even try to check your map $\mathbf Z[i] \rightarrow \mathbf Z[\sqrt{-5}]$ is a ring homomorphism? A homomorphism is not just symbol-pushing.
Apr
17
comment Restriction of topological ring isomorphism
Since you speak categorically, first one needs to know that $R^\times$ belongs to the category of topological groups!
Apr
17
comment Restriction of topological ring isomorphism
The previous comment focuses only on the topology, but the group structure alongside it is also important for your question to have meaning. Could you please tell us what you think the topology is on $R^\times$ that makes it a topological group?
Apr
17
comment prove that $x^2 + 5 =y^3$ has no solutions for $x,\ y \in \mathbb{Z}$
Do you know about ideal class groups? The ring $\mathbf Z[\sqrt{-5}]$ has class number $2$, which is relatively prime to the exponent $3$ in the equation, so even though $\mathbf Z[\sqrt{-5}]$ is not a UFD one can prove that $x+\sqrt{-5}$ must be a cube in $\mathbf Z[\sqrt{-5}]$, which is the kind of conclusion you'd expect if $\mathbf Z[\sqrt{-5}]$ were a UFD. Trying to solve $x + \sqrt{-5} = (a + b\sqrt{-5})^3$ in integers $a$ and $b$ leads to a contradiction by looking at the coefficient of $\sqrt{-5}$ on both sides. That's how you solve this with algebraic number theory.
Apr
17
comment prove that $x^2 + 5 =y^3$ has no solutions for $x,\ y \in \mathbb{Z}$
When you say $y = 1$ from $y^6 = 1$, actually you can only say $y = \pm 1$. But you still get a contradiction after that, as before.
Apr
17
comment An approach to proving that $\Bbb{Q}[x,y]/(x^3-y^2)$ is isomorphic to $\Bbb{Q}[t^2,t^3]$
It means nothing to "consider" them as separate variables. Evaluation at $(t^2,t^3)$ is a ring homomorphism $\mathbf Q[x,y] \rightarrow \mathbf Q[t^2,t^3]$ that is obviously surjective. Prove the kernel is $(x^3-y^2)$, i.e., for a polynomial $f(x,y)$ in $\mathbf Q[x,y]$ you have $f(t^2,t^3) = 0$ only if $f(x,y)$ is divisible by $x^3-y^2$. Note that your list of relations $x^{3n}=y^{2n}$ is woefully incomplete; those are just the easiest relations. They are definitely not all of them. The point of computing the kernel of that map is to show all relations are consequences of $x^3=y^2$.
Apr
15
comment Power series in p-adic integers
This is discussed in nearly every book on $p$-adic analysis. Have you read about convergence of power series in any book on $p$-adics (Gouvea, Koblitz, Robert, ...)?
Apr
15
comment When is the normality of field extensions transitive?
The field $\mathbf Q(\sqrt{2},\sqrt[3]{2},\zeta_3)$ is normal over $\mathbf Q$. It is a splitting field for $(x^2-2)(x^3-2)$ over $\mathbf Q$. You need to completely review this material, since you do not seem to have a strong base of examples that you understand well. If you are taking a class on Galois theory then have a talk with your instructor.
Apr
15
comment When is the normality of field extensions transitive?
There is no absolute concept of splitting field. The whole idea is relative. You speak about the splitting field of a polynomial over a given field. And any two splitting fields of a chosen polynomial over a given field are isomorphic to each other by an isomorphism fixing that given field. You have a mistaken idea of what a splitting field means.
Apr
15
comment Show that the Euclidean algorithm works for Gaussian integers.
Did you try anything towards solving this? And do you already know how to show the Euclidean algorithm works in simpler cases like $\mathbf Z$ and $K[x]$ ($K$ a field)? You need to understand the details of more basic cases before you try to explain harder cases.
Apr
15
comment When is the normality of field extensions transitive?
The extension $\mathbf Q(\sqrt{2},\zeta_3,\sqrt[3]{2})/\mathbf Q(\sqrt{2})$ is normal. The top field is a splitting field for $x^3 - 2$ over the bottom field. I don't think you're really going to get a satisfactory answer that is not just a translation of the condition into the language of Galois theory. Just for comparison, when is normality of subgroups transitive? There is no good universal answer to that.
Apr
9
comment Number of real embeddings $K\to\overline{\mathbb Q}$
The number of real embeddings is exactly the number of real roots of the minimal polynomial for $\sqrt{1+\sqrt{2}}$ over $\mathbf Q$.