Kernel of $k[a,b]to k[r^3,r^4], ;;f(a,b)mapsto f(r^3,r^4)$

Question

Let $k$ be a field, let $k[r^3,r^4]$ be a subalgebra of $k[r]$, and define the map
$$
varphi:k[a,b]to k[r^3,r^4],quad f(a,b)mapsto f(r^3,r^4).
$$
I want to show $ker(varphi)=(a^4-b^3)$. If $f(a,b)in (a^4-b^3)$, then $f(a,b)=g(a,b)cdot(a^4-b^3)$ for some $g(a,b)in k[a,b]$. Then $$f(r^3,r^4)=g(r^3,r^4)cdot((r^3)^4-(r^4)^3)=g(r^3,r^4)cdot(r^{12}-r^{12})=g(r^3,r^4)cdot0=0,$$
so $f(a,b)inker(varphi)$. I am stuck on the reverse direction. If I let $h(a,b)inker(varphi)$, then I know that $h(r^3,r^4)=0$. But where can I go from here? We need to show that $h(a,b)$ is a $k[a,b]$-multiple of $a^4-b^3$.

tkf · Accepted Answer

Given $alphainkervarphi$ we may subtract off some multiple of $(a^4-b^3)$ to attain an element of the form:$$p_0(a)+p_1(a)b+p_2(a)b^2,$$
where $p_0,p_1,p_2$ are polynomials over $k$.
We know that $$p_0(r^3)+p_1(r^3)r^4+p_2(r^3)r^8=0.$$
$p_0(r^3)$ is a $k$-linear combination of $r^i$ with $iequiv 0 mod 3$.
$p_1(r^3)r^4$ is a $k$-linear combination of $r^i$ with $iequiv 1 mod 3$.
$p_2(r^3)r^8$ is a $k$-linear combination of $r^i$ with $iequiv 2 mod 3$.
Thus no monomial $r^i$ occurs in more than one of the above.  However it cannot occur in just one as the sum of the three terms above is $0$.
We conclude that $p_0=p_1=p_2=0$.

Kernel of $k[a,b]to k[r^3,r^4], ;;f(a,b)mapsto f(r^3,r^4)$

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