Recent content by kmitza

I am learning some complex analysis as it is a prerequisite for the masters program that I was accepted into and I didn't take it yet during my bachelors. I am using some lecture notes in Slovene and I have run into a problem that has proven troublesome for me : If ##g: D \rightarrow \mathbb{C}...

Let ##S## be a set of n geometric objects in the plane. The intersection graph of ##S## is a graph on ##n## vertices that correspond to the objects in ##S##. Two vertices are connected by an edge if and only if the corresponding objects intersect. Show that the number of intersection graphs of...

Maybe I am mistaken but if I know the degree of the extension and I know it is Galois, don't I know that group is going to be of the order exactly the same as the degree? So I know that the order of the whole Galois group is p(p-1)? Now I know that I have a subgroup of order p which is...

As the summary says we have ## f(x) = x^n - \theta \in \mathbb{Q}[x] ##. We will call the pth primitive root ## \omega ## and we denote ##[\mathbb{Q}(\omega) : \mathbb{Q}] = j##. We want to show that the Galois group is generated by ##\sigma, \tau## such that $$ \sigma^j = \tau^p = 1...

Yes you are right about the second one, thanks for the swift answer. I am working on doing the first one and I will reply with the solution if I find one, I can say that trying a few examples seems to indicate that the result holds

We have Galois extension ## K \subset L ## and element ##\alpha \in L## and define polynomial $$f = \prod_{\sigma \in Gal(L/K)} (x - \sigma(\alpha))$$ Now we want to show that ## f \in K[x] ## which is relatively easy to see because we can take ##\phi(f)## for any ## \phi \in Gal(L/K) ## then...

Thanks for the tip!

I am going to give up a bit more on the given problem. We start with polynomial ## x^27 -x ## over GF(3)[x] and we factorize it using a well known theorem it turns out it factorises into the product of monic polynomials of degree 1 and 3, 11 of them all together. We then choose one of those...

Okay so immediately after writing my comment I think I figured it out : $$aba = b \implies ab = ba^{-1} \text{ and } ba = a^{-1}b$$ so now if I have a string of the form ##aabb## I can write $$ aabb + aba^{-1}b = b a^{-1}a^{-1}b = ba^{-1}ba = bbaa$$ and similar for arbitrary strings so in fact...

Um yes that is correct but it doesn't really help me, in general you can say that ##b^n(x,y) = (x+n,y) ## if n is odd and ##(x+n,1-y)## if n is even but without knowing the structure of the group elements I don't know how to calculate the whole orbit. I think that all members should be of the...

So before I start I technically do now that the group I am dealing with is just a representation of the Klein bottle but I am not supposed to use that as a fact because the goal of the problem is to derive that information. Problem: Let G be a group of with two generators a and b such that aba...

Oh my god that's so much simpler thank you for showing me, when it comes to the square it is an honest mistake I didn't see I added it

Um I am not sure what you mean I double checked with wolfram and hand just now and the one that has $$\pi + ei$$ as a root is $$h(x) = x^2 -2x\pi + e^2 + \pi^2$$ the other one has $$-\pi + ie$$ and it's conjugate as roots, right?

Yeah sorry my notation wasn't good, bad choice using a as a variable... I think I fixed it now

Hey thanks for saying hi, I submitted my attempt at one of the problems :)