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Math Help - Proving that an ideal is prime

  1. #1
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    Proving that an ideal is prime

    Let R= ℚ[X,Y] and let I=<X + 1>

    I am trying to prove that I is a prime ideal of R.

    I have tried arguing the following way (but i am unsure whether it is correct because the ring has two variables X,Y):

    Suppose that f,g ∈ℚ[X,Y] and that fg∈ <X + 1>

    That is, (x+1)|fg.

    Hence, fg(-1)=0, that is f(-1)g(-1)=0 .

    Therefore either f(-1)=0 , that is (X+1)|f, or g(-1)=0, that is (x+1)|g.

    Thus either f or g is in the ideal <X+1>.

    Therefore the ideal is prime.


    Is this a correct argument?

    How would i show that I is not a maximal ideal of R?

    Thanks
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  2. #2
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    Quote Originally Posted by Siknature View Post
    Let R= ℚ[X,Y] and let I=<X + 1>

    I am trying to prove that I is a prime ideal of R.

    I have tried arguing the following way (but i am unsure whether it is correct because the ring has two variables X,Y):

    Suppose that f,g ∈ℚ[X,Y] and that fg∈ <X + 1>

    That is, (x+1)|fg.

    Hence, fg(-1)=0, that is f(-1)g(-1)=0 .

    Therefore either f(-1)=0 , that is (X+1)|f, or g(-1)=0, that is (x+1)|g.

    Thus either f or g is in the ideal <X+1>.

    Therefore the ideal is prime.


    Is this a correct argument?

    How would i show that I is not a maximal ideal of R?

    Thanks

    The argument is correct though you must be sure that here applies the same as with one unknown...

    As for maximality: <x+1>\,<x+1,y> ...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post

    As for maximality: <x+1>\,<x+1,y> ...

    Tonio
    Are you saying that <x +1> is not maximal because <x+1,y> is an ideal which is greater than <x+1> ?
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  4. #4
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    Quote Originally Posted by Siknature View Post
    Are you saying that <x +1> is not maximal because <x+1,y> is an ideal which is greater than <x+1> ?

    Yes...somehow I ommited the "<" sign between those two ideals.

    Tonio
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