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Math Help - Ideal of the ring

  1. #1
    Junior Member Greg98's Avatar
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    Ideal of the ring

    Hello,
    Let's assume that \mathbb{R}^2 is a ring, which operations are addition and multiplication using components (eg. u + v = (u_1 + v_1, u_2 + v_2)). The task is to prove that x-axis and y-axis are ideals of this ring. Following theorem must be used in the proof:
    "Ring homomorphism's f:R \rightarrow S kernel Ker(f) is R's ideal".

    I think the problem isn't very hard, when the correct homomorphism is found. I tried some, but I couldn't discover anything that works. So, any help is welcome. Thank you very much!
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  2. #2
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    Well, the x-axis is, by definition, the set

    X=\{(a,0)\mid a\in \mathbb{R} \}.

    If you want to prove that X is an ideal using the statement you have written, we need a ring R and a ring homomorphism f:\mathbb{R}^2\rightarrow R such that X=\ker f.

    How about taking R=\mathbb{R}, and defining

    f:\mathbb{R}^2\rightarrow R
    (a,b)\mapsto a

    (projection onto the first coordinate)? It should be fairly simple to prove that f is indeed a ring homomorphism, and that the kernel of f is exactly the set of points located on the x-axis.

    You can do a completely analogous thing to show that the Y axis is also an ideal.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by topspin1617 View Post
    Well, the x-axis is, by definition, the set

    X=\{(a,0)\mid a\in \mathbb{R} \}.

    If you want to prove that X is an ideal using the statement you have written, we need a ring R and a ring homomorphism f:\mathbb{R}^2\rightarrow R such that X=\ker f.

    How about taking R=\mathbb{R}, and defining

    f:\mathbb{R}^2\rightarrow R
    (a,b)\mapsto a

    (projection onto the first coordinate)? It should be fairly simple to prove that f is indeed a ring homomorphism, and that the kernel of f is exactly the set of points located on the x-axis.

    You can do a completely analogous thing to show that the Y axis is also an ideal.
    I think topspin made a slight clerical error. To show that X\unlhd \mathbb{R} one should consider the second canonical projection a,b)\to b" alt="\pi_2:\mathbb{R}^2\to\mathbb{R}a,b)\to b" /> since the kernel of this is clearly \mathbb{R}\times\{0\}=X.
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