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Math Help - Ring Properties

  1. #1
    Member Maccaman's Avatar
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    Ring Properties

    Determine if the given set (below) is a ring. If it is a ring, state if the ring is commutative, has identity, is an integral domain, and is a field.

    (1)The set  S = \lbrace \ \begin{pmatrix} a&b\\b&a\end{pmatrix} \  | a,b,\in \ \mathbb{R} \rbrace under normal matrix addition and multiplication.

    (2) The set of  \mathcal{F} of all functions  f: \mathbb{R} \ \rightarrow \ \mathbb{R} under pointwise addition (as the addition operation) and composition ( as the multiplication operation).

    Thanks to anyone who can help
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  2. #2
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    Quote Originally Posted by Maccaman View Post
    Determine if the given set (below) is a ring. If it is a ring, state if the ring is commutative, has identity, is an integral domain, and is a field.

    (1)The set  S = \lbrace \ \begin{pmatrix} a&b\\b&a\end{pmatrix} \ | a,b,\in \ \mathbb{R} \rbrace under normal matrix addition and multiplication.

    (2) The set of  \mathcal{F} of all functions  f: \mathbb{R} \ \rightarrow \ \mathbb{R} under pointwise addition (as the addition operation) and composition ( as the multiplication operation).

    Thanks to anyone who can help
    see my next post!
    Last edited by NonCommAlg; April 20th 2009 at 03:24 PM.
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  3. #3
    Super Member Gamma's Avatar
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    multiply

    1) it is pretty easy to check the ring conditions, i mean it clearly is a group under addition it is just basically component wise if you think of it as in \mathbb{R}^4. The multiplication too is easy to check just multiply by another matrix in that form, make sure it is closed and then multiply on the other side by that matrix and see if you get the same thing. Something tells me if you are capable of TeXing up the matrices you are capable of multiplying out two 2x2 matrices. I just did it really quick and it looks like its a commutative ring.

    2) pointwise addition is pretty easy to check is a group, the question is about composition. As long as the functions are defined over all of \mathbb{R} you should be fine. Certainly the composition of two functions from \mathbb{R}\rightarrow \mathbb{R} will still be a function \mathbb{R}\rightarrow \mathbb{R}.

    I am just a little concerned about like if you had your first function as f(x)=-e^x and then composed it with g(x)=ln(x) like this function would not be defined anywhere, but I assume when they defined the family of functions it has to be defined on all of \mathbb{R}. To see that is is definitely not commutative try f(x)=x^2 and g(x)=1+2x. Hope this is of some help.

    Ah I see NCA already posted, well instead of deleting it ill let you read mine too maybe it might help as well. And I think NCA might check the commutativity of 1) it looks commutative to me.
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  4. #4
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    Quote Originally Posted by Gamma View Post
    1) it is pretty easy to check the ring conditions, i mean it clearly is a group under addition it is just basically component wise if you think of it as in \mathbb{R}^4. The multiplication too is easy to check just multiply by another matrix in that form, make sure it is closed and then multiply on the other side by that matrix and see if you get the same thing. Something tells me if you are capable of TeXing up the matrices you are capable of multiplying out two 2x2 matrices. I just did it really quick and it looks like its a commutative ring.

    2) pointwise addition is pretty easy to check is a group, the question is about composition. As long as the functions are defined over all of \mathbb{R} you should be fine. Certainly the composition of two functions from \mathbb{R}\rightarrow \mathbb{R} will still be a function \mathbb{R}\rightarrow \mathbb{R}.

    I am just a little concerned about like if you had your first function as f(x)=-e^x and then composed it with g(x)=ln(x) like this function would not be defined anywhere, but I assume when they defined the family of functions it has to be defined on all of \mathbb{R}. To see that is is definitely not commutative try f(x)=x^2 and g(x)=1+2x. Hope this is of some help.

    Ah I see NCA already posted, well instead of deleting it ill let you read mine too maybe it might help as well. And I think NCA might check the commutativity of 1) it looks commutative to me.
    Gamma is absolutely right about 1). i actually thought it was the set of all 2 by 2 matrices! haha anyway, 1) is a commutative ring with identity but it's not an integral domain because

    for example: \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}=0, and thus it cannot be a field. the second one is also a ring with identity but it's not commutative. it's also not a domain because it has 0 divisors.

    for example define f(1)=2 and f(x)=0, \ x \neq 1. then fof=0. so 2) is not a division ring (and thus not a field).
    Last edited by NonCommAlg; April 20th 2009 at 04:10 PM.
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