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Thread: Algrbra Help!!! rings and ideals :(

  1. #1
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    Algrbra Help!!! rings and ideals :(

    I need some help:

    Prove that if an ideal I of R contains a unit then I=R

    Prove that a commutative ring is a field iff it it contains exactly 2 ideals

    Prove that if the ring R is commutative and local then R*=R\I

    some hints/tips would be great.

    Thanks
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by gloriad View Post
    I need some help:

    Prove that if an ideal I of R contains a unit then I=R
    Let $\displaystyle u \in I$, where u is a unit.
    Then there is $\displaystyle u^{-1} \in R$ such that $\displaystyle u^{-1}u =1$. Since, I is an ideal, $\displaystyle 1 \in I$
    Let $\displaystyle r \in R$. Then $\displaystyle r \cot 1 \in I$. Thus $\displaystyle R \subseteq I$. Therefore, I = R..
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by gloriad View Post
    I need some help:



    Prove that a commutative ring is a field iff it it contains exactly 2 ideals
    Let F be a field.
    If $\displaystyle A = \left\{ 0 \right\}$, then it is ok. Suppose $\displaystyle A \not = \left\{ 0\right\}$ be an ideal of F. Then A contains a unit. From the previous, A = F.

    Lemma: Let R be a commutative ring with unity. and M a maximal ideal of R. Then R/M is a field iff M is maximal.(You can show this.. )

    Suppose, F contains exactly 2 ideals. Since $\displaystyle \left\{ 0 \right\}$ is the only proper ideal of F, then $\displaystyle \left\{ 0 \right\}$ is a maximal ideal, and so $\displaystyle F/\left\{ 0 \right\}$ is a field which is isomorphic to F. Thus, F is a field. QED.
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  4. #4
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    Let R be a commutative ring (with unity) that contains only two ideals, {0} and R. Let I be an ideal of R. If R is a ring such that $\displaystyle 1\not = 0$ (kalagota forgot to mention that) then choose $\displaystyle a\not = 0$. Now define $\displaystyle \left< a\right> = \{ ra|r\in R\}$. Then clearly this is an ideal of R and is non-trivial so it means $\displaystyle \left< a \right> = R$ which means there exists a $\displaystyle b$ so that $\displaystyle ab=1$. This shows all the non-zero elements of R has inverses, this makes R a field. Q.E.D.
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  5. #5
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    can you help me with the third part of the question?

    can you help me with this part:

    Prove that if the ring R is commutative and local then R*(units of R)=R\I, where I is the largest proper ideal of R?

    it seems to me that it follows from the last two parts of the question but i can't phrase it correctly.

    Thanks
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let R be a commutative ring (with unity) that contains only two ideals, {0} and R. Let I be an ideal of R. If R is a ring such that $\displaystyle 1\not = 0$ (kalagota forgot to mention that) then choose $\displaystyle a\not = 0$. Now define $\displaystyle \left< a\right> = \{ ra|r\in R\}$. Then clearly this is an ideal of R and is non-trivial so it means $\displaystyle \left< a \right> = R$ which means there exists a $\displaystyle b$ so that $\displaystyle ab=1$. This shows all the non-zero elements of R has inverses, this makes R a field. Q.E.D.
    maybe, or maybe not (did i forget). It's because i used to prove the exercise using the lemma i stated.. anyways, thanks TPH!!
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