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Thread: Substitution Homomorphism from F[x,y] to F[t]

  1. #1
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    Substitution Homomorphism from F[x,y] to F[t]

    Hello everyone. I have been struggling with the following problem from "Algebra An Approach via Module Theory" by Adkins and Weintraub.

    Let F be a field and let f be a homomorphism from F to F. Extend f to a homomorphism from F[x,y] to F[t] by setting f(x) = t^2, f(y) = t^3. Then the kernel of f is the principal ideal generated by y^2 - x^3.

    Its clear that the ideal generated by y^2 - x^3 is contained in the kernel of f. To show the opposite inclusion I have resorted to writing out a general polynomial in the variables x and y and hoping to factor it (using the assumption it is in the kernel of f) so that y^2 - x^3 is a factor. This has so far been very tedious and does not seem to be the best way to go with this problem. But I am having trouble finding inspiration on this one.
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  2. #2
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    Re: Substitution Homomorphism from F[x,y] to F[t]

    Quote Originally Posted by Kopeck View Post
    Hello everyone. I have been struggling with the following problem from "Algebra An Approach via Module Theory" by Adkins and Weintraub.

    Let F be a field and let f be a homomorphism from F to F. Extend f to a homomorphism from F[x,y] to F[t] by setting f(x) = t^2, f(y) = t^3. Then the kernel of f is the principal ideal generated by y^2 - x^3.

    Its clear that the ideal generated by y^2 - x^3 is contained in the kernel of f. To show the opposite inclusion I have resorted to writing out a general polynomial in the variables x and y and hoping to factor it (using the assumption it is in the kernel of f) so that y^2 - x^3 is a factor. This has so far been very tedious and does not seem to be the best way to go with this problem. But I am having trouble finding inspiration on this one.
    let $\displaystyle I = \langle y^2-x^3 \rangle$ and $\displaystyle u \in \ker f$. then, since $\displaystyle y^2 \equiv x^3 \mod I,$ we have $\displaystyle u \equiv p(x) + yq(x) \mod I$, for some $\displaystyle p(x), q(x) \in F[x]$. since $\displaystyle I \subseteq \ker f$ and $\displaystyle u \in \ker f,$ we have $\displaystyle p(x) + yq(x) \in \ker f,$ i.e. $\displaystyle p(t^2)+t^3q(t^2)=0$. but all powers of $\displaystyle t$ in $\displaystyle p(t^2)$ are even numbers and all powers of $\displaystyle t$ in $\displaystyle t^3q(t^2)$ are odd numbers. thus $\displaystyle p(t^2)+t^3q(t^2)=0$ implies $\displaystyle p = q = 0$ and so $\displaystyle u \equiv 0 \mod I,$ i.e. $\displaystyle u \in I. \ \Box$
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  3. #3
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    Re: Substitution Homomorphism from F[x,y] to F[t]

    Thank you very much for the help NonCommAlg. This is a very cool technique.
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