Recent content by RVP91

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    HomA(M,HomA(N,K)) is isomorphic to HomA(N,HomA(M,K))

    Homework Statement Let A be a commutative ring with identity element. Prove that HomA(M,HomA(N,K)) is isomorphic to HomA(N,HomA(M,K)). Homework Equations The Attempt at a Solution I believe it is best to start by defining a map, f: HomA(M,HomA(N,K) → HomA(N,HomA(M,K)) for ψ: M...
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    (Z/10557Z)* as Abelian Groups using Chinese Remainder Theorem

    If I was to try to work this out I would use the Chinese Remainder Theorem and since 10557 = 3^3 . 17 . 23 end up with (Z/10557Z)* isomorphic to (Z/27Z)* x (Z/17Z)* x (Z/23Z)* isomorphic to C18 x C16 x C22 where Cn represents the Cyclic group order n. How would I then write this as Cn1 x Cn2...
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    Can (Z/10557Z)* be written as Cn1 x Cn2 x Cn3 with n1 dividing n2 dividing n3?

    If I was to try to work this out I would use the Chinese Remainder Theorem and since 10557 = 3^3 . 17 . 23 end up with (Z/10557Z)* isomorphic to (Z/27Z)* x (Z/17Z)* x (Z/23Z)* isomorphic to C18 x C16 x C22 where Cn represents the Cyclic group order n. How would I then write this as Cn1 x Cn2...
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    Probability branching process proof

    Oh right. I'm really confused now. Is there any chance you could perhaps give me the first few lines of the proof and then some hints on how to continue please?
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    Probability branching process proof

    So normally would it be zero order as the offspring at each stage are independent of the any offspring around them in the same generation. "By conditioning on the value of X1, and then thinking of future generations as a particular generation of the separate branching processes spawned by...
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    Probability branching process proof

    In particular could someone explain what it is saying when it says "By conditioning on the value of X1, and then thinking of future generations as a particular generation of the separate branching processes spawned by these children" I think this is essentially the key but I don't understand...
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    Probability branching process proof

    Could you explain further, after reconsideration I know for sure my original working was totally incorrect. Could anyone help me out? Possibly start me off? Thanks.
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    Probability branching process proof

    By conditioning on the value of X1, and then thinking of future generations as a particular generation of the separate branching processes spawned by these children, show that Fn(s), defined by Fn(s) = E(s^Xn), satisfies Fn(s) = F(Fn−1(s)) ∀n ≥ 2. I need to prove the above result and...
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    How Do You Solve Vector Space Dimension Challenges?

    So is that all correct now? Also thank you so much for all the help!
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    How Do You Solve Vector Space Dimension Challenges?

    Would this work? A={(a,b,c,d,0,e,f,g,h,0)} B={(a,0,b,c,d,e,f,g,h,i)} Then A and B span R^10 since a linear combiantion Q(a,b,c,d,0,e,f,g,h,0) + P(a,0,b,c,d,e,f,g,h,i) = (Qa+Qb,Qb,Qc+Pb,...) = (x1,x2,x3,x4,x5,x6,x7,x8,x9,x10). And also A∩B i think is equal to {(a,0,b,c,0,d,e,f,g,0). Is that...
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    How Do You Solve Vector Space Dimension Challenges?

    I'm not getting very far, could you give another hint perhaps? Thanks.
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    How Do You Solve Vector Space Dimension Challenges?

    For the answer of 7 would the following be okay? A={(a,b,0,c,d,e,f,g,0,h) |a,b,c,d,e,f,g,h are Real} B={(a,b,0,c,d,e,f,g,h,i) |a,b,c,d,e,f,g,h,i are Real} Then dim A = 8, dim B = 9, but A∩B = ({a,b,0,c,d,e,f,g,0,0)|a,b,c,d,e,f,g are Real} and this has dimension 7? Also going back to my earlier...
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    How Do You Solve Vector Space Dimension Challenges?

    Does it not need to equal 7? As my answer before was dim(A∩B) = 7 or 8, not 8 or 9?
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    How Do You Solve Vector Space Dimension Challenges?

    Would it equal to dim(A) and so 8?
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