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Math Help - Maximal Ideals in Integral Domain

  1. #1
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    Maximal Ideals in Integral Domain

    Let R be an integral domain with quotient field K = \{ \frac {a}{b} | a,b \in R, b \neq 0 \} , and let I and J be ideals of R. Let X denote the set of maximal ideals of R.

    Let t \in K . Prove that if t \in \bigcap _{M \in X } R_M , then t \in R

    pf.

    Now, R_M = \{ \frac {r}{s} | r,s \in R \ , \ s \in R  \ not \ M \} , so t = \frac {a}{b}, a \in R, b \in (R \ not \ M)

    To be honest, I'm quite lost after this, how should I go on with this problem? thanks.
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  2. #2
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    What if t \in K-R?

    \exists (a,b) \in R\times (R-\{0\})\ ;\ t=\frac{a}{b} , and b is non invertible (or t=ab^{-1} \in R) and does not divide a.

    So there is a maximal ideal M_{0} in R such that b \in M_{0} (Krull's theorem). Thus t isn't an element of \bigcap _{M \in X } R_M, because it doesn't belong to R_{M_{0}}.
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  3. #3
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    Quote Originally Posted by clic-clac View Post
    What if t \in K-R?

    \exists (a,b) \in R\times (R-\{0\})\ ;\ t=\frac{a}{b} , and b is non invertible (or t=ab^{-1} \in R) and does not divide a.

    So there is a maximal ideal M_{0} in R such that b \in M_{0} (Krull's theorem). Thus t isn't an element of \bigcap _{M \in X } R_M, because it doesn't belong to R_{M_{0}}.
    unfortunately the conclusion is not correct! you can have both b \in M_0 and \frac{a}{b} \in R_{M_0}. for example \frac{2}{6} \in \mathbb{Z}_{2\mathbb{Z}}. i will show that we actually have \bigcap_{\mathfrak{m} \in X} R_{\mathfrak{m}}=R: define \mathfrak{a}=\{r \in R: \ rt \in R \}.

    obviously \mathfrak{a} is an ideal of R. if \mathfrak{a}=R, then 1 \in \mathfrak{a}, and hence t \in R. if \mathfrak{a} \neq R, then there exists a maximal ideal \mathfrak{m}_0 \supseteq \mathfrak{a}. now since t \in R_{\mathfrak{m}_0}, there exist a \in R, \ b \in R-\mathfrak{m}_0 such that t=\frac{a}{b}.

    thus bt=a \in R. hence b \in \mathfrak{a} \subseteq \mathfrak{m}_0, which is impossible since b \in R- \mathfrak{m}_0. this proves that \bigcap_{\mathfrak{m} \in X} R_{\mathfrak{m}} \subseteq R. also note that R \subseteq R_{\mathfrak{m}} for all maximal ideals \mathfrak{m}, because the map \varphi_{\mathfrak{m}}: R \longrightarrow R_{\mathfrak{m}}

    defined by \varphi_{\mathfrak{m}}(r)=\frac{r}{1} is injective. thus R \subseteq \bigcap_{\mathfrak{m} \in X} R_{\mathfrak{m}}. hence: R=\bigcap_{\mathfrak{m} \in X} R_{\mathfrak{m}}. \ \ \Box
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    Thanks for your help!!!

    So in this case, we would also have I = \bigcap _{M \in X } IR_M then?

    Now one way of this direction should be automatic, as  \bigcap _{M \in X } IR_M is contained in I.

    But is it true for the other way?
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    Quote Originally Posted by tttcomrader View Post

    So in this case, we would also have I = \bigcap _{M \in X } IR_M then? yes!
    since I \subseteq IR_{\mathfrak{m}}, \ \forall \mathfrak{m} \in X, we have: I \subseteq \bigcap_{\mathfrak{m} \in X} IR_{\mathfrak{m}}. for the other side, the idea is the same as above: let t \in \bigcap_{\mathfrak{m} \in X} IR_{\mathfrak{m}}. define \mathfrak{a}=\{r \in R: \ rt \in I \}. clearly \mathfrak{a} is an ideal of R. now if \mathfrak{a}=R,

    then 1 \in \mathfrak{a} and hence t \in I. if \mathfrak{a} \neq R, then there exists \mathfrak{m}_0 \in X: \ \mathfrak{a} \subseteq \mathfrak{m}_0. now since t \in IR_{\mathfrak{m}_0}, there exist a \in I, \ b \in R-\mathfrak{m}_0 : \ t=\frac{a}{b}. thus bt=a \in I, i.e. b \in \mathfrak{a} \subseteq \mathfrak{m}_0. contradiction! \Box
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  6. #6
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    Thanks for your massive help!!!

    My last question here is: Prove that I=J if and only if IR_M=JR_M, for each maximal ideal M of R.

    Proof so far.

    Suppose that I=J, then IR_M = I = J = JR_M, since I and J are ideals.

    Conversely, suppose that IR_M = JR_M, pick an element in IR_M, then it has the form i \frac {a}{b}, a \in R, b \in R \ not \ M, i \in I .

    Now, I=J implies that there exists an element j \in J such that j=i, thus i \frac {a}{b} = j \frac {a}{b} \in JR_M, so we have IR_M \subseteq JR_M, and the reverse is the same.

    Is this correct? My solution look a little bit too easy compare to the other problems...

    Thanks!!!
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  7. #7
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    Quote Originally Posted by tttcomrader View Post
    Thanks for your massive help!!!

    My last question here is: Prove that I=J if and only if IR_M=JR_M, for each maximal ideal M of R.

    Proof so far.

    Suppose that I=J, then IR_M = I = J = JR_M, since I and J are ideals.

    Conversely, suppose that IR_M = JR_M, pick an element in IR_M, then it has the form i \frac {a}{b}, a \in R, b \in R \ not \ M, i \in I .

    Now, I=J implies that there exists an element j \in J such that j=i, thus i \frac {a}{b} = j \frac {a}{b} \in JR_M, so we have IR_M \subseteq JR_M, and the reverse is the same.

    Is this correct? My solution look a little bit too easy compare to the other problems...

    Thanks!!!
    no! your proof is not correct: it's not true that I=IR_{\mathfrak{m}}. if R is an integral domain, then we can only say that I \subseteq IR_{\mathfrak{m}}. but we may have I \neq IR_{\mathfrak{m}}.

    for example if R=\mathbb{Z}, \ I=3\mathbb{Z}, \ \mathfrak{m}=2\mathbb{Z}, then I \subsetneq IR_{\mathfrak{m}}=\{\frac{3k}{2n+1}: \ k, n \in \mathbb{Z} \}. anyway, here's how you should solve the problem:

    it's obvious that I=J implies that IR_{\mathfrak{m}}=JR_{\mathfrak{m}} because by definition: IR_{\mathfrak{m}}=\{\frac{a}{s}: \ a \in I, \ s \in R-\mathfrak{m} \}=\{\frac{a}{s}: \ a \in J, \ s \in R-\mathfrak{m} \}=JR_{\mathfrak{m}}.

    conversely suppose IR_{\mathfrak{m}}=JR_{\mathfrak{m}}, for all maximal ideals \mathfrak{m} of R. we'll show that I \subseteq J: let a \in I and define \mathfrak{L}=\{r \in R: \ ra \in J \}. see that \mathfrak{L} is an

    ideal of R. if \mathfrak{L}=R, then 1 \in \mathfrak{L} and thus a \in J. otherwise, \mathfrak{L} \subseteq \mathfrak{m}, for some maximal ideal \mathfrak{m}. now since a \in I \subseteq IR_{\mathfrak{m}}=JR_{\mathfrak{m}}, we have a=\frac{b}{s}, for some

    b \in J, \ s \in R - \mathfrak{m}. but then sa=b \in J, and thus s \in \mathfrak{L} \subseteq \mathfrak{m}, which is impossible because s \in R-\mathfrak{m}. this completes the proof of I \subseteq J. an exactly the

    same argument shows that J \subseteq I and therefore I=J. \ \Box
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