Recent content by Kiwi1

How about this problem: add all of the numbers 1,2,3,...,1000. You could take it at face value and add them sequentially You could re-order them to get 1000+(1+999) + (2+998) + ... and sum them up that way You could re-order them in the same way but recognise how the series will terminate to...

Once we start to speculate on what a typo is things get hard. Perhaps they mean ONE factor is \(\leq 2\). The example I have shown has one factor of degree \(\leq 2\) and one \(\geq 2\). Looking at the subsequent questions it would make more sense to me that they would want me to prove ONE...

OK so solved Q1. It was straightforward to show (by contradiction) that w is in the center of K. From there it is easy enough to show that \(F(d,\omega)\) is the root field. But now I can't solve Q2. In fact I can falsify the assertion of Q2 as follows: Let F be the field of rational...

Any suggestions on how I should approach question C1? Every time I think I have a solution I find that I have made the implicit assumption either that F is abelian or that the roots of w are in the center of F. I don't think either assumption is valid. If I let K be the root field of the poly...

Solved it! Suppose \(I_2\) is a normal extension of \(I_1\). Then there exists a(x) such that: \(I_2 \cong \frac{I_1[x]}{<a(x)>} \supseteq I_1\). Now this isomorphism fixes \(I_1\) so by Theorem (i) the isomorphism is in fact an automorphism. So by problem 4: \(I_2^* \cong...

I have answered Q1-Q3. I am unsure of my answer for Q4 and am stuck on Q5. Notation: \(I^*=Gal(K:I)\). That is the subgroup of G=Gal(K:F) that fixes I. Q4. I2 is conjugate to I1. iff \(\exists i \in G:I_2=i(I_1)\) iff \( I^*_2=[i(I_1)]^*=iI_1^*i^{-1}\) by Q3. iff \( I^*_2=iI_1^*i^{-1}\) iff...

Thanks guys. F(a,b)* is the field of automorphisms of, a larger field containing F(a,b), that don't change each element in F(a,b). With that knowledge solving the problem is straightforward.

Could it mean the subset of elements that have an inverse?

G. Shorter questions relating to automorphisms and Galois Groups Let F be a field, and K a finite extension of F. Suppose \(a,b \in K\). Prove parts 1-3. 2. \(F(a,b)^*=F(a)^* \cap F(b)^*\) Surely, they mean the union of F(a) and F(b) and not the intersection? There is no reason to think...

I have been working my way through the book "A book of Abstract Algebra" by Charles C. Pinter. By the time I get to the end it will have taken me 2 years of self study, solving every exercise during my daily commute. A great little book! The last two chapters are titled: Chapter 32: Galois...

I think I solved it. I have "Theorem 7: Let K be the root field of some polynomial q(x) over F. For every irreducible polynomial p(x) in F[x], if p(x) has one root in K, then p(x) must have all of its roots in K." In my problem I have \(q(x) = (x-d_2)(x+d_2)=x^2-d_2^2=x^2-d_1+\frac a2 \in...

Prove that if \(p(x)=x^4+ax^2+b\) is irreducible in F[x], then \(F[x]/<p(x)>\) is the root field of p(x) over F. My Attempt: 1. Let F(c) = \(F[x]/<p(x)>\) where c is a root of p(x). Then F(c) is a degree 4 extension over F because c is the root of a 4th order irreducible polynomial in F[x]...

Thanks, I think I have found it. I was working on an HTML file on my local hard drive. Sometimes I use Mathjax off the web and sometimes from a copy on my HDD. It turns out I get the bars when I use the HDD copy. So I downloaded a fresh copy of MathJax and so far it seems to be problem...

I was using Chrome v49. I just upgraded to Chrome v50. I have tried opening my document with IE11 and that was fine. In Chrome (either version) if I open my file then all is fine, when I scroll to the bottom of the doc and back to the top the bars appear.

When I enter math like \(x^2\) I often get a vertical bar shown to the right of my math when I print the document or view it on screen. Something like this: \(x^2|\) I get this both on this site and also when I write HTML using mathjax. When I preview this post I don't get the vertical bar...