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Thread: Condition for Commutative Ring to be a Field.

  1. #1
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    Condition for Commutative Ring to be a Field.

    Hi,

    Suppose $\displaystyle R \neq \{0\}$ is a commutative ring with no zero divisors.
    Suppose that for every $\displaystyle a \in R$, $\displaystyle a \neq 0$, we have $\displaystyle aR = R$.

    Prove R is a field.

    I am not sure what to do... just need to show that each $\displaystyle a \in R $ has an inverse (ie, each element is a unit)? Not sure how to go about it.

    Thanks!
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  2. #2
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    I suppose we know that $\displaystyle <a>\ = R\ , \forall a \in R$...
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  3. #3
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    I think I made some progress? not sure if it is valid...

    Since $\displaystyle <a>\ = R$, then we can write each element $\displaystyle r \in R$ as $\displaystyle r = ac$ for some $\displaystyle c \in R$.

    But since $\displaystyle a$ itself is in $\displaystyle R$, then $\displaystyle ac = a$ for some $\displaystyle c \in r$.

    Since $\displaystyle R$ is commutative, $\displaystyle ac = ca = a$
    Does this mean that we have a unity? How do we prove that $\displaystyle c$ works for every element of $\displaystyle R$?

    Once we can show we have a unity, I think the rest is done, because:
    if $\displaystyle r \in R$, then $\displaystyle <r> = R$. So each element of $\displaystyle R$ can be written as $\displaystyle rd$ for some $\displaystyle d \in R$. Since $\displaystyle 1_R \in R$, $\displaystyle 1_R = rd$ for some $\displaystyle d \in R$. Since $\displaystyle R$ is commutative, $\displaystyle 1_R = rd = dr$, so by the definition of unique inverses, each element has an inverse, so each element is a unit, so $\displaystyle R$ is a field.


    Can anyone look over this argument and verify it/spot a problem? I'd be very grateful, thanks!
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  4. #4
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    Quote Originally Posted by matt.qmar View Post
    I think I made some progress? not sure if it is valid...

    Since $\displaystyle <a>\ = R$, then we can write each element $\displaystyle r \in R$ as $\displaystyle r = ac$ for some $\displaystyle c \in R$.

    But since $\displaystyle a$ itself is in $\displaystyle R$, then $\displaystyle ac = a$ for some $\displaystyle c \in r$.

    Since $\displaystyle R$ is commutative, $\displaystyle ac = ca = a$
    Does this mean that we have a unity? How do we prove that $\displaystyle c$ works for every element of $\displaystyle R$?

    Once we can show we have a unity, I think the rest is done, because:
    if $\displaystyle r \in R$, then $\displaystyle <r> = R$. So each element of $\displaystyle R$ can be written as $\displaystyle rd$ for some $\displaystyle d \in R$. Since $\displaystyle 1_R \in R$, $\displaystyle 1_R = rd$ for some $\displaystyle d \in R$. Since $\displaystyle R$ is commutative, $\displaystyle 1_R = rd = dr$, so by the definition of unique inverses, each element has an inverse, so each element is a unit, so $\displaystyle R$ is a field.


    Can anyone look over this argument and verify it/spot a problem? I'd be very grateful, thanks!

    1) $\displaystyle aR=R\Longrightarrow \exists\,c_a\in R\,\,s.t.\,\,ac_a=a$

    2) $\displaystyle aR=R\Longrightarrow \forall\,x\in R\,\,\exists\, r_x\in R\,\,s.t.\,\,ar_x=x$

    3) $\displaystyle \forall\,x=ar_x\in R\,,\,\,xc_a=(ar_x)c_a=r_x(ac_a)=r_xa=x\Longrighta rrow c_a$ is the

    whole ring's multiplicative unit.

    Tonio
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