Interesting thing: there are two definitions of the Riemann Stieltjes integral, which are not equivalent, but are equivalent for the plain Riemann integral.
showing no ring containing Z16 has roots of that polynimial is equivalent to showing a characteristic 16 ring containing a root of that polynomial is impossible
one person's basic facts are another person's alien magic - MO quote
Hi, off topic question: How is $Gal(\mathbb{Q} (2^{1/3},\zeta_3): \mathbb{Q}(2^{1/3}))$ a subgroup of $Gal(\mathbb{Q} (2^{1/3},\zeta_3): \mathbb{Q})$ when the former has 4 elements. Doesn't that contradict Lagrange?
hey sometimes the prof sucks bar none k. Look what's happening to my calc iv class...if he doesn't curve heavily, everyone is going to get an F. I AM NOT KIDDING
calc is a sophomore level class...doesn't count towards the ba in Math so wtf am I in there? well if it wasn't for that class I would've been in trouble
@cap why do you think the former has 4 elements? it's a degree 2 extension, and the only automorphism is complex conjugation. in general if L/M/K is a tower then G(L/M) is a subgroup of G(L/K): every automorphism of L that fixes M pointwise must in particular fix K pointwise
so if any a**hole like Ted wants to be a total B**** and says s*** like I don't deserve to be a Math Major, well f*** their negative crap. I've been through this twice already and prove those jealous haters wrong.
Let $p(x)$ be irreducible in $F [x]$. If $[f (x)] ̸= [0]$ in $F [x]/⟨p(x)⟩$ and $h(x) ∈ F [x]$ prove that there is a $g(x) ∈ F [x]$ such that $[f (x)][g(x)] = [h(x)]$ in $F [x]/⟨p(x)⟩.$
Ted Shifrin Sigh. This whole thing is depressing me beyond belief. I'm trying to be in a good mood. So I'm out of here. Bye, y'all. http://chat.stackexchange.com/transcript/36/2013/12/5/22-23 ahhh so it happened three days ago
That's what I keep on saying to myself...it's not a competition yet students boast too much about their 100s in an exam. ughhhhhh makes everyone feel bad. Of course I passed my exams, but what the hell man I used to take them without getting nervous. I scored like B's and A's during my community college years...what happened?
Well, get off math.SE. Look at your calc iv book, open it. Hate it for the first 5 minutes. DO NOT LOSE FOCUS and do not close the book!! Then you shall read the first page. and then the second. you will have some thoughts telling you how boring it is. Then you will ignore those thoughts, perhaps you will think about skipping over the material. but you won't and you will think harder for about 10 minutes. and you will continue until you are hungry, or tired. then you can eat and sleep!
@Pedro: Not everyone is as smart as you. So for us mere mortals that don't have the purely mathematical mindset, we get curves so our egos aren't completely destroyed.
@DonLarynx I'm not judging or whatever, maybe getting 100/120 is nearly impossible, I don't know how the exam works. Maybe you're asked to do a part, like Ted did in his final. You can explain...!
It was my first time looking at advanced calculus, my second semester in mathematics. @Pedro, in order to not drop out of the major, my teacher curves because this material is not easy - and he knows it.
I think passing with a $B$ in advanced calculus given that criteria is superior.
Don, his problem seems to be a little less about math and a little more about MSE. This is the post I saw earlier today that made me remember him: "Ping the moderators and you will get the following reply: "We are watching what we can to see if there is anything we can see that the script has not. Keep us informed.". But no action will be taken and they will keep watching till the entire site collapses."
Then why do we have sucky moderators? This is what happens when we have big government, some people expect others to do all the work - and then in the end, there is an overload of the system and everything collapses. Kind of like in "The machine stops".
@Eric: I will be done with algebra and I will be a free man forever, in 7 hours. Meanwhile I am looking forward to analysis final exam. Isn't that ironic?
That was fun... I finished showing $\mathbb Q(x)/\mathbb Z[x]$ is not an injective $\mathbb Z[x]$-module even though it is an injective $\mathbb Z$-module--even a divisible $\mathbb Z[x]$-module.