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Math Help - Several problems regarding Ideals

  1. #1
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    Several problems regarding Ideals

    Hello, i am stumped on these problems, and i am wondering if anybody could help me

    1. Prove or disprove that in the ring Z[X], <2X> = <2,X>

    2. Suppose that I is an ideal of Q[X] which contains both X^{2} + 2X + 4 and X^{3} - 3. Show that I = Q[X].

    3. Prove that in the ring Z[X], <2>U<X> is not an ideal
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  2. #2
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    Quote Originally Posted by maroon_tiger View Post
    1. Prove or disprove that in the ring Z[X], <2X> = <2,X>
    To show that \left< 2x \right> = \left< 2,x \right> we need to show \left< 2x \right> \subseteq \left<2 ,x\right> and \left<2,x\right> \subseteq \left<2x\right>.

    2. Suppose that I is an ideal of Q[X] which contains both X^{2} + 2X + 4 and X^{3} - 3. Show that I = Q[X].
    Note x^3 - 3 = (x-2)(x^2+2x+4) + 5 this means \gcd (x^3 - 3,x^2+2x+4) = \gcd (x^2+2x+4, 5) = 1 by Euclidean algorithm. Thus, by relative primeness there exists f(x),g(x)\in \mathbb{Q}[x] such that f(x)(x^2+2x+4) + g(x) (x^3-3) = 1. Since I is an ideal it means f(x)(x^2+2x+4) + g(x) (x^3-3) = 1 \in I. All ideals which contain 1 have to be the improper ideal, i.e. I = \mathbb{Q}.

    3. Prove that in the ring Z[X], <2>U<X> is not an ideal
    Find two elements a,b\in \left< 2\right> \cup \left< x \right> so that a+b\not \in \left<2\right> \cup \left< x \right>.
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  3. #3
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    thanks for the help
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    To show that \left< 2x \right> = \left< 2,x \right> we need to show \left< 2x \right> \subseteq \left<2 ,x\right> and \left<2,x\right> \subseteq \left<2x\right>.
    i tried that but i believe that they arent equal b/c x^{3} + 2X^{2} + 2 has a factor of <2,X> b/c of 2(X^{2} +1) + X(X^{2}) but x^{3} + 2X^{2} + 2 doesnt have a factor of <2x> unless im wrong


    also, on another problem,

    1. Prove that in the ring Z[X], <2> /\ <X> = <2x>
    i understand why this would be true. would i prove this by the properties of an ideal? by assuming that <2> = {2f(x), x in some ideal A} and <X> = {Xg(y). some y in B}, then a point "p" in <2>^<X> such that {p exists in 2f(x)Xg(x)} and prove that 2f(x)Xg(x) = 2xf(x)? is that correct?
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