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

1. ## 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.

2. ## Re: Substitution Homomorphism from F[x,y] to F[t]

Originally Posted by Kopeck
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 $I = \langle y^2-x^3 \rangle$ and $u \in \ker f$. then, since $y^2 \equiv x^3 \mod I,$ we have $u \equiv p(x) + yq(x) \mod I$, for some $p(x), q(x) \in F[x]$. since $I \subseteq \ker f$ and $u \in \ker f,$ we have $p(x) + yq(x) \in \ker f,$ i.e. $p(t^2)+t^3q(t^2)=0$. but all powers of $t$ in $p(t^2)$ are even numbers and all powers of $t$ in $t^3q(t^2)$ are odd numbers. thus $p(t^2)+t^3q(t^2)=0$ implies $p = q = 0$ and so $u \equiv 0 \mod I,$ i.e. $u \in I. \ \Box$

3. ## Re: Substitution Homomorphism from F[x,y] to F[t]

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