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Math Help - Condition for Commutative Ring to be a Field.

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

    Hi,

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

    Prove R is a field.

    I am not sure what to do... just need to show that each 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 <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 <a>\ = R, then we can write each element r \in R as r = ac for some c \in R.

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

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

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

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

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

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


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

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

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

    3) \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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