quick question: does the irreducibility of a polynomial in a field depend on whether there are roots in it, or whether you can write it as the product of 2 polynomials with coefficients in the root, because I want to know if $x^4-2$ is irreducible in Q($2^{1/2}$), and I know there aren't roots in it, but can't you also write it as ($x^2 + $2^{1/2}$)($x^2 - $2^{1/2}$)
I know that there are no roots to $x^4 - 2$ in $\Bbb{Q}[\sqrt{2}]$, but does that mean my factorization is an invalid one?: $(x^{2}+\sqrt{2})((x^{2}-\sqrt{2}$
@DHMO having written my matrix, it's clear that the characteristic polynomial of that matrix is a polynomial with $\sqrt 2 + \sqrt 3$ as a solution. I haven't thought much about how linear algebra might tell you it's the minimal such
@Faust: When I videoed all my multivariable math lectures, I discovered that not having a class there to interact naturally with totally ruins the experience/lecture. (One of the videos did not download correctly and I had to re-record the lecture, but with no students.)
Interesting, @Faust. I've never thought about that. We'd need an isometry of the Poincaré disk taking any three boundary points to any three boundary points. Can we find one of those?
Suppose that there are two ideal quadrilaterals A and B, and that A can be decomposed into ideal triangles a_1, a_2, and B can be decomposed into ideal triangles b_1, b_2. We know that There is an isometry c_1 which maps a_1 to b_1, and an isometry c_2 which maps a_2 to b_2. why do we know that c_1 and c_2 are different?
they came up a few times in the probability course, but since we didn't know what a convolution is the professor just handwaved a bit and glossed over the details
@Ted In any case my argument works for operators of the form $p(D)$, $p$ a polynomial with positive coefficients, only even powers, and nonzero constant term.
@Daminark You mean for peoples learning very basic maths ? It might be a little bit too hard to grasp the depth of what angles really are for highschool maths
And @Astyx it did some stuff about the function defining the area under a circle and said that $\cos(x)$ was the unique number such that $A(x) = \frac{\cos(x)}{2}$ or something
Yeah, it works but I dunno, it feels somewhat contrived
Like, it's neither the power series/exponential version that everyone rolls with, nor is it the right triangle trigonometry which at least is easier to get your mind around, you know?
Hi, @DHMO, yes, it works that way, too. I'm using the Open Mapping Theorem for a proof, that's why I asked about $f(U)$
The statement to be proven is as follows:
Let $f$ be analytic on a non-empty domain, $D$. If $f^2(z)= \overline {f(z)}$ for all $z \in D$, then prove that $f^3$ is constant on D and deduce that $f$ is also constant on $D$.
My proof goes this way: suppose that $f^3$ is non-constant, then by the Open Mapping Theorem, if $U \subseteq D$ is an open set, then $f^3(U)$ is also an open set
@Evinda, are you familiar with the rank-nullity theorem? If the Kernel is zero, then its nullity is zero. Hence, the matrix is non-singular and the solution to $Ax=b$ is unique.
@Evinda, if the Kernel is not zero, then its nullity is not zero. By the rank-nullity theorem, the sum of the rank and the nullity is equal to the number of columns. But if there are $N$ columns, then the rank of that matrix is less than $N$, meaning there are zero rows. Hence, for a non-homogenous equation, there are more unkowns than equations. Hence, the solution won't be unique.
Also, the nullity of a matrix represents the number of zero rows.
@Evinda, the nullity of the matrix is equal to the dimension of the null space. Every matrix can be expressed as a linear transformation and vice versa. Let's say you have a square matrix $A$. Then its nullity is the number of zero rows in its row echelon form (or reduced, if you want it).
When the kernel contains only the null vector, then the dimension of the kernel, which is also its nullity, is zero. By the rank-nullity theorem, the square matrix is full rank, and is thus non-singular. Thus, if it is a coefficient matrix, then the solution set is unique.
@Evinda what goes wrong if you just do basically the same row operations as Gaussian elimination and just let it end early with some unknowns that parameterize your solution set?
Problem: If H and K are subgroups of G, then |H v K : H | \ge |K : H \cap K|, where H v K = <H \cup K > denotes the join of H and K. Fact: If H and K are subgroups of G, then |K : H \cap K | \le |G : H|. Am I correct in thinking that the problem would follow trivially from preceding fact, since H and K are subgroups of H v K?
Let $X$ be an odd prime subset of the integers, that is $\pm $ odd primes $P_o$. Define $D(X) = \{ \pm(x - p) : p\geq x, \ x \in X, p\in P_o\}$. Then $D(\cup_i X_i) = \cap_i D( X_i)$ for arbitrary unions and intersections.
Proof: $x - y \in D(\cup_i X_i) \iff y \geq x', \ \forall x' \in \cu...
Let $X$ be an odd prime subset of the integers, that is $\pm $ odd primes $P_o$. Define $D(X) = \{ \pm(x - p) : p\geq x, \ x \in X, p\in P_o\}$. Then $D(\cup_i X_i) = \cap_i D( X_i)$ for arbitrary unions and intersections.
Proof: $x - y \in D(\cup_i X_i) \iff y \geq x', \ \forall x' \in \cu...
