I want to show that $K[t^2,t^3]_{(t^2,t^3)}$ is not regular, and to do so I want to show that $dim_F(m/m^2)=2$ for $m=(t^2,t^3).K[z^2,t^3]_{(t^2,t^3)}=(\frac{t^2}{1},\frac{t^3}{1})\subseteq K[t^2,t^3]_{(t^2,t^3)}$ and $F=K[t^2,t^3]_{(t^2,t^3)}/m$, which proves the statement since $dimK[t^2,t^3]_{(t^2,t^3)}=1$.
I show that $\{[\frac{t^2}{1}],[\frac{t^3}{1}]\}$ is a basis of $m/m^2$ and I need your help to tell me if what I did is correct, and maybe guide me on how to show that it's free. We have that $[\frac{t^i}{1}]=0+m^2$ $\forall i \geq 4$ since $m^2=(\frac{t^4}{1},\frac{t^5}{1})$.
We take $[\frac{p(t^2,t^3)}{q(t^2,t^3)}]\in m/m^2$ with $\frac{p(t^2,t^3)}{q(t^2,t^3)}\in m=(t^2,t^3).K[t^2,t^3]_{(t^2,t^3)}$. So $p(t^2,t^3)\in (t^2,t^3)\subseteq K[t^2,t^3]$.
We can write $$p(t^2,t^3)=f(t^2,t^3)t^2+g(t^2,t^3) \implies[\frac{p(t^2,t^3)}{q(t^2,t^3)}]=(\frac {f(t^2,t^3)}{q(t^2,t^3)}+m).[\frac{t^2}{1}]+(\frac{g(t^2,t^3)}{q(t^2,t^3)}+m).[\frac{t^3}{1}]\in span \{[\frac{t^2}{1}],[\frac{t^3}{1}]\}$$ So $\{[\frac{t^2}{1}],[\frac{t^3}{1}]\}$ generates $m/m^2$. Now we show that it's free. Suppose it's not, then $\exists (\frac{p(t^2,t^3)}{q(t^2,t^3)}+m) \in F=K[t^2,t^3]_{(t^2,t^3)}/m$ such that $[\frac{t^2}{1}]=(\frac{p(t^2,t^3)}{q(t^2,t^3)}+m).[\frac{t^3}{1}]$ (or vice versa). We have $[\frac{t^2}{1}]=[\frac{p(t^2,t^3)t^3}{q(t^2,t^3)}]$ but if $p(t^2,t^3)$ contains non constant terms they will vanish since they will be of degree 4 or more since they're multiplied by $t^3$. So $[\frac{t^2}{1}]=[\frac{at^3}{q(t^2,t^3)}] \implies \frac{t^2q(t^2,t^3)-at^3}{q(t^2,t^3)}=\frac{g(t^2,t^3)t^4}{h(t^2,t^3)}+\frac{u(t^2,t^3)t^5}{v(t^2,t^3)}$ which doesn't lead to much.
I want to argue on the degree but there are too many polynomials. Is it possible to show the linear independence with this ?