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asked If $G$ acts on $A$ faithfully/freely/transitively, what can we say about its action on $\hat A$?
Aug
25
revised What's the “real” reason a finite map has finite fibers?
edited body
Aug
23
comment Divisors of zero in polynomial ring
To add some texture to Fredrik Meyer's answer and Thomas Andrews' comment: to feel why $x^7-x$ should not be considered to be the zero polynomial even though it evaluates to zero on every element of $\mathbb{F}_7$, consider that this same polynomial is also defined over every ring containing $\mathbb{F}_7$, on which it no longer necessarily defines the zero function. In fact, if $L$ is a field containing $\mathbb{F}_7$ properly, then $x^7-x$ will be nonzero on $L$ except on $\mathbb{F}_7$.
Aug
23
answered Non-split chain complex which is chain-homotopy equivalent to its homology sequence
Aug
22
accepted If a chain complex is homotopy equivalent to its homology, is it split?
Aug
22
revised If a chain complex is homotopy equivalent to its homology, is it split?
acknowledged peter a g's comments
Aug
22
comment If a chain complex is homotopy equivalent to its homology, is it split?
@peterag - might you make your comments an answer to give me something to accept? (I still find it hard to believe that this is something Weibel made a mistake about!)
Aug
22
comment If a chain complex is homotopy equivalent to its homology, is it split?
@Dorebell - and if $y\in\operatorname{im}(1-\tau)$ i.e. $y=(x-\tau x)$ then $\tau y = \tau x - \tau^2 x = 0$, so $1-\tau$ maps into $\ker\tau$. (Because it acts as the identity on it, it must be onto.) For the direct sum decomposition, $x = \tau(x) + (1-\tau)(x)$ gives any $x\in A$ as the sum of $\tau(x)\in \operatorname{im}\tau, (1-\tau)(x)\in\ker \tau$, and the representation is unique since $\ker\tau \cap \operatorname{im}\tau = 0$ since $\tau$ acts trivially on the former and as the identity on the latter.
Aug
22
comment If a chain complex is homotopy equivalent to its homology, is it split?
@Dorebell - I added an addendum that gives the argument without relying on this decomposition, but in any case I'm sure of it. $\pi$ acts as the identity on any element of $A_n$, so it surjects. General claim: if $A$ is a module (over any ring) and $\tau\in\operatorname{End}A$ is a projection (i.e. $\tau^2=\tau$), then $1-\tau$ is a projection to $\tau$'s kernel, and $A = \ker\tau \oplus \operatorname{im}\tau$. Proof: $(1-\tau)^2 = 1-2\tau + \tau = 1-\tau$, so $1-\tau$ is a projection; if $x\in\ker \tau$, then $(1-\tau)(x) = x$, so $1-\tau$ acts as the identity on $\ker \tau$; cont'd...
Aug
22
comment If a chain complex is homotopy equivalent to its homology, is it split?
@peterag - yes it is p. 18 line 3! Thank you!
Aug
22
revised If a chain complex is homotopy equivalent to its homology, is it split?
removed requirement that the ring $R$ be commutative and unital; added cleaner proof attempt
Aug
22
asked If a chain complex is homotopy equivalent to its homology, is it split?
Aug
9
comment how did Cardano obtain three solutions for cubic?
Also, use the math-history tag to clarify the historical intent of the question to answerers.
Aug
9
comment how did Cardano obtain three solutions for cubic?
Aside: I see you are new to this site. The culture here is very one-clear-question-per-post. So if you feel that your question about Cardano has been answered, but still want to know more about Euler, you should mark it answered, and ask a new question about Euler, containing a link to this question to show the context.
Aug
9
revised how did Cardano obtain three solutions for cubic?
added a math-history tag
Aug
9
comment how did Cardano obtain three solutions for cubic?
I haven't read Euler's work on cubics so I can't give you a definitive answer. This essay ms.uky.edu/~sohum/ma330/files/eqns_4.pdf seems to confirm that Euler is responsible for the idea of using cube roots of unity to obtain all 3 solutions. (I have read Cardano's Ars Magna which is why I felt empowered to answer your question.)
Aug
9
answered how did Cardano obtain three solutions for cubic?
Aug
9
comment When does a representation of $H\subset G$ on $V$ extend to a representation of $G$ on $V$?
@DerekHolt - thanks, just the kind of thing I was looking for.
Aug
7
comment When does a representation of $H\subset G$ on $V$ extend to a representation of $G$ on $V$?
Do you know if there is any literature on how to identify the special case where a representation of $H$ extends to $G$?
Aug
7
accepted When does a representation of $H\subset G$ on $V$ extend to a representation of $G$ on $V$?