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  1. #1
    Senior Member Sampras's Avatar
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    field

    1. Prove that  R = \{a+bi: a,b \in \mathbb{Z} \} is a subring of  \mathbb{C} and that  M = \{a+bi: 3|a \ \text{and} \ 3|b \} is a maximal ideal in  R .


    Showing it's a subring is obvious. Just show closure under addition and multiplication and that  0_{\mathbb{C}} \in R . Also show that the solution of the equation  a+x = 0_{\mathbb{C}} is in  R .


    To show that  M is maximal consider the following:  M \subset J \subset R means that there is some  j \in J such that  j \notin M where  j = r+si . So  3 \not| \ |j| = r^2+s^2 = (r+si)(r-si) . And any ideal that contains  r+si must contain  1 so that  J = R .

    Is this correct?




    2. To show that  R/M is a field we know this from the fact that  M is maximal. How do you show it has 9 elements?
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  2. #2
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    Quote Originally Posted by Sampras View Post
    1. Prove that  R = \{a+bi: a,b \in \mathbb{Z} \} is a subring of  \mathbb{C} and that  M = \{a+bi: 3|a \ \text{and} \ 3|b \} is a maximal ideal in  R .


    Showing it's a subring is obvious. Just show closure under addition and multiplication and that  0_{\mathbb{C}} \in R . Also show that the solution of the equation  a+x = 0_{\mathbb{C}} is in  R .


    To show that  M is maximal consider the following:  M \subset J \subset R means that there is some  j \in J such that  j \notin M where  j = r+si . So  3 \not| \ |j| = r^2+s^2 = (r+si)(r-si) . And any ideal that contains  r+si must contain  1 so that  J = R .

    Is this correct?


    No! Why would J = R if J contains r + si = j??? J = R iff 1 is in J, and this is what you must show.



    2. To show that  R/M is a field we know this from the fact that  M is maximal. How do you show it has 9 elements?

    Try to find 9 different elements in the quotient ring. It isn't that hard as you may think at first, and it'll give some hours of leisure...and learning.

    Tonio
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  3. #3
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    Quote Originally Posted by Sampras View Post
    To show that  M is maximal consider the following:  M \subset J \subset R means that there is some  j \in J such that  j \notin M where  j = r+si . So  3 \not| \ |j| = r^2+s^2 = (r+si)(r-si) . And any ideal that contains \color{red}r+si must contain \color{red}1 so that  J = R .
    Why is this true?
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  4. #4
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    Quote Originally Posted by Sampras View Post
    1. Prove that  R = \{a+bi: a,b \in \mathbb{Z} \} is a subring of  \mathbb{C} and that  M = \{a+bi: 3|a \ \text{and} \ 3|b \} is a maximal ideal in  R .


    Showing it's a subring is obvious. Just show closure under addition and multiplication and that  0_{\mathbb{C}} \in R . Also show that the solution of the equation  a+x = 0_{\mathbb{C}} is in  R .


    To show that  M is maximal consider the following:  M \subset J \subset R means that there is some  j \in J such that  j \notin M where  j = r+si . So  3 \not| \ |j| = r^2+s^2 = (r+si)(r-si) . And any ideal that contains  r+si must contain  1 so that  J = R .

    Is this correct?

    Not as far as I can see. The element j= r+si is an arbitrary element in J; how does containing this element makes an ideal the whole ring?!
    I think you must show that the norm of such an element j is coprime with the norm of any element in M and thus...

    Read here, in particular 10.2: it is exactly what you need and want for this and the next question:
    http://www.fen.bilkent.edu.tr/~franz/nt/ch10.pdf

    Tonio






    2. To show that  R/M is a field we know this from the fact that  M is maximal. How do you show it has 9 elements?
    .
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  5. #5
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    in the following i'll use \mathbb{F}_3 for the field of order 3 instead of \mathbb{Z}/3\mathbb{Z}:

    the ideal \mathcal{N}=<x^2+1> is a maximal ideal of \mathbb{F}_3[x] because x^2+1 is irreducible. thus \frac{\mathbb{F}_3[x]}{\mathcal{N}} is a field of order 9. define f: R \longrightarrow \frac{\mathbb{F}_3[x]}{\mathcal{N}}. by f(a+bi)=a+bx + \mathcal{N}.

    see that f is a surjective ring homomorphism and clearly: \ker f = \{a+bi : \ 3 \mid a, \ 3 \mid b \}=\mathcal{M}. therefore \frac{R}{\mathcal{M}} \cong \frac{\mathbb{F}_3[x]}{\mathcal{N}} and you're done.
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  6. #6
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    Quote Originally Posted by Sampras View Post
    1. Prove that  R = \{a+bi: a,b \in \mathbb{Z} \} is a subring of  \mathbb{C} and that  M = \{a+bi: 3|a \ \text{and} \ 3|b \} is a maximal ideal in  R .


    Showing it's a subring is obvious. Just show closure under addition and multiplication and that  0_{\mathbb{C}} \in R . Also show that the solution of the equation  a+x = 0_{\mathbb{C}} is in  R .


    To show that  M is maximal consider the following:  M \subset J \subset R means that there is some  j \in J such that  j \notin M where  j = r+si . So  3 \not| \ |j| = r^2+s^2 = (r+si)(r-si) . And any ideal that contains  r+si must contain  1 so that  J = R .

    Is this correct?




    2. To show that  R/M is a field we know this from the fact that  M is maximal. How do you show it has 9 elements?
    This is another approach.

    It is clear that every odd prime in \mathbb{Z} has a form either 1(mod 4) or 3(mod 4) and an even prime 2=(1+i)(1-i) is reducible in R=\mathbb{Z}[i]. If p \equiv \text{1(mod 4)}, then p=x^2 + y^2 with x and y integers by Fermat's two squares theorem. Further, p=x^2 + y^2=(x + yi)(x - yi) and neither x+yi nor x-yi is a unit because their norms are bigger than 1. Thus p \equiv \text{1(mod 4)} is reducible.

    Since R=\mathbb{Z}[i] is an integral domain (not a field) and a PID, we have p \equiv \text{3(mod 4)} as irreducible elements and <p>= \{a+bi: p|a \ \text{and} \ p|b \} are maximal ideals in R=\mathbb{Z}[i] (For example, <3>, <7>, <11>, <19>, ..., are maximal ideals in R=\mathbb{Z}[i]).
    Last edited by aliceinwonderland; October 29th 2009 at 03:49 AM.
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