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Thread: an example from Gallian, [ring theory]

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    an example from Gallian, [ring theory]

    an example from Gallian, [ring theory]-screenshot_2.png
    Can you please explain to me this problem(if possible, in details, sorry )? I dont know why (1,0) is there, why there isn't a unity...
    any help appreciated
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    Re: an example from Gallian, [ring theory]

    Quote Originally Posted by lampageu View Post
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    Can you please explain to me this problem. I don't know why (1,0) is there, why there isn't a unity...
    It would be helpful if you tell us the name of the textbook you are using. Is this for an introduction to algebra or is it for a course in ring theory?

    As to $(1,0)$, that is just descriptive condition on the members in the set forming the ring. You may want to start with seeing how point is relevant to the two operations.
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    Re: an example from Gallian, [ring theory]

    The point (1, 0) is there because that is what the problem says! Do you know what a "ring" is? Do you know what a "unity" is?

    A ring is a set of objects with two operations, called "addition" and "multiplication" defined such that
    a) Addition is "associative"- that is, a+(b+ c)= (a+ b)+ c for all a, b, c in the ring.
    b) Addition is "commutative"- a+ b= b+ a for all a, b in the ring.
    c) There exist an "additive identity"- a member of the group, usually called "0" such that a+ 0= a.
    d) Every member of the set has an "additive inverse"- for every a in the set, there exist b such that a+ b= 0.
    e) Multiplication is "associative"- a(bc)= (ab)c for all a, b, c in the ring.
    f) Multiplication "distributes" over addition- a(b+ c)= ab+ ac for all a, b, c in the ring.

    Multiplication is not necessarily commutative in a ring. And saying there isn't a "unity" means there is no "multiplicative identity", no member "1" such that a "times" 1= a. And, of course, if there is no multiplicative identity, there can be no "multiplicative inverses".

    Here the set is defined as the set of all functions, f, g, h, etc. such that f(1)= 0, g(1)= 0, h(2)= 0. Addition and multiplication are defined "pointwise"- that is, the function, f+ g, applied to x is f(x)+ g(x) and the function, fg, applied to x is f(x)g(x) where, since f and g are "real valued", f(x)+ g(x) is the sum of two real numbers and f(x)g(x) is the product of two real numbers.

    To prove this is ring you need to verify all of the conditions above. For example, a: if f, g, and h are in the set then f+ (g+ h) if, by definition, the function such that f+ (g+ h), applied to x, is f(x)+ (g(x)+ h(x)). Since these are "real valued functions", f(x), g(x), and h(x) are all real numbers and addition of real numbers is associative: f(x)+ (g(x)+ h(x))= (f(x)+ g(x))+ h(x) which is the same as the function (f+ g)+ h applied to x: f+ (g+ h)= (f+ g)+ h. Similarly for b: f+ g is the function that, applied to x, gives f(x)+ g(x). Again, f(x) and g(x) are real numbers. Since addition of real numbers is commutative, f(x)+ g(x)= g(x)+ f(x) which is the same as the function f+ g applied to x: f+ g= g+ f. One also needs the fact that the sum of two continuous functions, and the product of two continuous functions, is continuous.

    If there were a "unity", a multiplicative identity, u, in the set then it would have to have the property that uf= f for all functions, f, in the set. In particular, if f(x)= x- 1, which is in the set, then (uf)(x)= (x+ 1)u(x)= f(x)= x+ 1 for all x. But, as long as x is not -1, we can divide by x+ 1 to get u(x)= 1 which does not satisfy "u(1)= 0".
    Last edited by HallsofIvy; Feb 9th 2017 at 05:40 AM.
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