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  • 20 votes cast
Jun
21
comment How does one construct the Galois field extension $GF((2^2)^3)$?
Thank you for answering. I have a question, why is it that $-1 = 1$, when I was doing the calculations by hand yesterday, I was using $-3=1$, and I had $\beta^3=3 \alpha \beta^2 + 3\alpha \beta + 3 \alpha$. Also, how do you find the inverse of $\alpha$, and then how exactly does $\beta^4= \beta^2 + \beta + \alpha^2$? If you can help with the relations of the elements in the coefficient field, then I think I'll be able to do the calculations.
Jun
20
comment How does one construct the Galois field extension $GF((2^2)^3)$?
@JyrkiLahtonen I don't think we're being asked to check if the polynomial is indeed primitive. I don't know very much yet, but I'm thinking the problem is to write down/find explicitly every element of the finite field as a linear combination of the basis elements of the corresponding cyclic group when, say, $\beta$ is the primitive element. I tried doing that and after very tedious computations wasn't able to show that $\beta^{63}=1$.
Jun
20
comment How does one construct the Galois field extension $GF((2^2)^3)$?
@ZevChonoles Yes.
Jun
20
comment How does one construct the Galois field extension $GF((2^2)^3)$?
@PatrickDaSilva I was typing the problem as was written, but yes what you wrote is what is meant.
Jun
7
comment How do I show that the following is a basis for the weak topology on $X$?
@BrianM.Scott Yes, that's what I meant. Apologies, I was typing from my phone and wasn't careful.
Jun
7
comment How do I show that the following is a basis for the weak topology on $X$?
@ThomasE. The definition I'm using is that the topology on $X$ defined by the semi norm $\{pf : f \in X^*\}$ is the weak topology.
May
3
comment How do I solve this PDE (diffussion equation) using the sepration of variables method?
Can I ask where is this taken from so that I may look it up in the library here?
Sep
23
comment How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
@Timbuc I didn't. And thanks for your help.
Sep
23
comment How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
@Timbuc No, only that $|Z| < 1$
Sep
23
comment How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
So, it's a mistake in the text?
Sep
23
comment How do I show $|\frac{i\overline{z}}{2} - \frac{i}{2}|=|z - 1|?$
Of course I tried that before asking here. This isn't my homework, btw.
May
17
comment Connected spaces: is there a mistake in the example below from James Munkres' Topology?
Thank you for your answer.
May
17
comment Connected spaces: is there a mistake in the example below from James Munkres' Topology?
Yes, you're right. I was skipping over the word 'contains'. And, no it doesn't say that the subspace is connected. Lots of confusion in my head, is all. Thanks for your help.
Oct
22
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
Right. So thanks again for all your help. Cheers. :)
Oct
22
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
I see we're talking just about the ordinals. Only when $|A \times A|=|A|$ for all infinite sets then that imply AC. And I know if it's a theorem of ZF that makes it a theorem of ZFC also. I didn't phrase my comment very well. What I meant to say earlier was that your first theorem looked to be that of ZFC and not ZF.
Oct
22
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
By Hessenberg sums for ordinals do you mean the canonical well-ordering of ordinal pairs? Used to prove that $\aleph_\alpha \times \aleph_\alpha = \aleph_\alpha$? And does not $|\kappa \times \kappa|=|\kappa|$ imply AC? Your first theorem seems also to be that of ZFC to me. Correct me if I'm wrong.
Oct
21
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
Thank you so very much for your answer.
Oct
21
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
@bof Yes, assuming AC.
Oct
21
comment How do I show that $\kappa^+ \le 2^\kappa$ for every cardinal $\kappa$?
@bof Yes I know Cantor's theorem. And all the cardinals are comparable only when you assume AC. I haven't seen a proof of $\lambda < \kappa^+$ implies $\lambda \le \kappa$. I'll look into that. I think it's best to assume I know everything, and answer me. I'd ask if I didn't understand something.
May
6
comment How do I prove that an uncountable subset $S$ of $\mathbb{R}$ has the in-between property?
@BrianM.Scott: I see your point. I didn't thought about that either.