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Math Help - Ring without maximal ideal

  1. #1
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    Ring without maximal ideal

    Hey there,

    Let R be the ring consisting of the power series of the form

    \sum_{n=1}^{\infty}a_{n}x^{\alpha_n} where a_n \in \mathbb{R} and 0 < \alpha_1 < \alpha_2 < \cdots are real numbers so that \lim_{x\to\infty}\alpha_n = \infty.

    I want to show that R is an integral domain without a maximal ideal. But I don't see how to start...
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  2. #2
    Senior Member Shanks's Avatar
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    If it has I as a maximal ideal, then R/I is a field.
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    And where is the contradiction?
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    Quote Originally Posted by EinStone View Post
    Hey there,

    Let R be the ring consisting of the power series of the form

    \sum_{n=1}^{\infty}a_{n}x^{\alpha_n} where a_n \in \mathbb{R} and 0 < \alpha_1 < \alpha_2 < \cdots are real numbers so that \lim_{x\to\infty}\alpha_n = \infty.

    I want to show that R is an integral domain without a maximal ideal. But I don't see how to start...
    for f=\sum_{n=1}^{\infty} a_n x^{\alpha_n} \in R with 0 < \alpha_1 < \alpha_2 < \cdots and a_1 \neq 0 define \deg f = \alpha_1. clearly \deg (f \cdot g)=\deg f + \deg g and so R is an integral domain (without the identity element!)

    now suppose that I is a maximal ideal of R. let S=\mathbb{R} + R and choose g \in R - I. let J=Sg. see that J is an ideal of R. since \forall h \in J: \ \deg h \geq \deg g, we have J \neq R.

    so J is a "proper ideal" of R. it's clear that I \neq J because g \in J - I. let f \in I. if f = 0, then clearly f \in J. so we'll assume that f \neq 0. suppose that \deg g > \deg f. then \frac{g}{f} \in R
    and therefore g \in Rf \subseteq I, which is a false result. so we must have \deg g \leq \deg f and so \frac{f}{g} \in S. thus f \in Sg = J. hence I \subset J \subset R, which contradicts maximality of I. \ \Box
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    I got 2 questions:
    1. How do you know that f, g, h have their a_1 \neq 0 (why is degree defined?)

    2. I dont understand the notation, what does S = \mathbb{R} + R and what does Sg or Rf mean?
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  6. #6
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    Quote Originally Posted by EinStone View Post
    I got 2 questions:
    1. How do you know that f, g, h have their a_1 \neq 0 (why is degree defined?)
    if a_j is the first non-zero element in the sequence \{a_1, a_2, \cdots \}, then \alpha_j (and not a_j) is called the degree of f =\sum a_n x^{\alpha_n}. of course we may always "assume" that a_1 \neq 0 and so \deg f = \alpha_1.


    2. I dont understand the notation, what does S = \mathbb{R} + R and what does Sg or Rf mean?
    the notations are standard and i don't know how you didn't understand them?? \mathbb{R} is the set of real numbers and S=\mathbb{R} + R = \{r + f: \ \ r \in \mathbb{R}, \ f \in R \}.

    also Sg = \{sg: \ s \in S \} and Rf = \{hf: \ h \in R \}.
    Last edited by NonCommAlg; December 6th 2009 at 07:31 AM.
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  7. #7
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    Ok thanks for clarification.

    Quote Originally Posted by NonCommAlg View Post
    the degree of f =\sum a_n x^{\alpha_n} \in R is the first non-zero element in the sequence \{\alpha_1, \alpha_2, \cdots \}.
    Here \{\alpha_1, \alpha_2, \cdots \} you mean \{a_1, a_2, \cdots \} right?


    Ok and another question:
    If we allow \alpha_1 = 0 then R has an identity and therefore a maximal ideal. Prove that in this case the maximal ideal is unique and identify it.

    I cant find the maximal ideal, anyone has an idea?
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  8. #8
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    Quote Originally Posted by EinStone View Post

    Here \{\alpha_1, \alpha_2, \cdots \} you mean \{a_1, a_2, \cdots \} right?
    that was a terrible typo i made. i just fixed it and you should take a look at it again.


    If we allow \alpha_1 = 0 then R has an identity and therefore a maximal ideal. Prove that in this case the maximal ideal is unique and identify it.
    I cant find the maximal ideal, anyone has an idea?
    well, if you allow \alpha_1=0, then you'll exactly get the ring S=\mathbb{R}+R, which i already mentioned. in this case R (your original ring) would be the unique maximal ideal of this new ring S.

    first see that R is indeed an ideal of S. now let I be any ideal of S not contained in R. so I has an element of the form f=a_1 + \sum_{j=2}^{\infty}a_jx^{\alpha_j}, \ \ a_1 \neq 0, \ 0 < \alpha_2 < \alpha_3 < \cdots. but then f
    would have an inverse in S and, since I is an ideal, we get 1=f \cdot \frac{1}{f} \in I. thus I=S. so every proper ideal of S is contained in R and hence R is the unique maximal ideal of S.
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  9. #9
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    Ok good I get this. To see that R is an ideal in S, it is sufficient to show that r*s \in R if r \in R and s \in S right?
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