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Math Help - Local ring

  1. #1
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    Local ring

    Hi,

    Could You help me with this task?

    "Proof that a commutative ring a is a local ring iff for any a, b \in A, from this that a+b=1 we have a is an invertible element or b is an invertible element."

    Proof. (\Rightarrow) a,b\in A - any element which satysfied equality a+b=1 and suppose that a,b\notin U(A), where U(A) is a set of invertible elements of A. We have then that a,b\in \mathfrak{m}. Since \mathfrak{m} is an ideal, then also 1=a+b\in \mathfrak{m}, and so \mathfrak{m}=A. This is contradiction, so a \in U(A) or b \in U(A).

    (\Leftarrow) - I'm not sure. Could You help me with this implication?
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  2. #2
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    Quote Originally Posted by Arczi1984 View Post
    Hi,

    Could You help me with this task?

    "Proof that a commutative ring a is a local ring iff for any a, b \in A, from this that a+b=1 we have a is an invertible element or b is an invertible element."

    Proof. (\Rightarrow) a,b\in A - any element which satysfied equality a+b=1 and suppose that a,b\notin U(A), where U(A) is a set of invertible elements of A. We have then that a,b\in \mathfrak{m}. Since \mathfrak{m} is an ideal, then also 1=a+b\in \mathfrak{m}, and so \mathfrak{m}=A. This is contradiction, so a \in U(A) or b \in U(A).

    (\Leftarrow) - I'm not sure. Could You help me with this implication?


    Let \displaystyle{S:=A-U(A) , and take x,y\in S . If x+y\notin S then

    x+y\in U(A)\Longrightarrow \exists c\in A\,\,s.t.\,\,1=c(x+y)=cx+cy\Longrightarrow cx\in U(A)\,\,or\,\,cy\in U(A) ,

    but in both cases we get a straightforward contradiction (can you see why?) , so it must be

    that x+y\in S and from here S is an ideal, and we're thus done.

    Tonio
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  3. #3
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    I found longer solution to this implication. Your is shorter and easier
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