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Thread: 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 $\displaystyle a, b \in A$, from this that $\displaystyle a+b=1$ we have $\displaystyle a$ is an invertible element or $\displaystyle b$ is an invertible element."

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

    $\displaystyle (\Leftarrow)$ - 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 $\displaystyle a, b \in A$, from this that $\displaystyle a+b=1$ we have $\displaystyle a$ is an invertible element or $\displaystyle b$ is an invertible element."

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

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


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

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

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

    that $\displaystyle x+y\in S$ and from here $\displaystyle S$ 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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