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Use the graded ring homomorphismwhere source degree $m$ maps to target degree $3m$. The homogeneous ideal of the twisted cubic is $\ker \phi$: a homogeneous polynomial vanishes on the parametrized curve precisely when its substitution vanishes identically. Here $k$ is infinite, being an algebraically closed field, so vanishing at all parameter points implies the polynomial identity.

Let $J$ be generated byTheir substitutions vanish, hence $J\subseteq \ker \phi$. To prove the reverse inclusion as an equality of ideals, we need more than a set-theoretic description of the zero locus.

Modulo $J$, replace $x_1^2$ by $x_0{x}_2$, $x_2^2$ by $x_1{x}_3$, and $x_1{x}_2$ by $x_0{x}_3$. Each replacement strictly decreases the sum of the exponents of $x_1$ and $x_2$, so the reduction terminates. It shows that the degree-$m$ quotient is spanned by the standard monomials of the twisted cubicFor $m\ge 1$ there are $(m+1)+m+m=3m+1$ of these. For $m=0$ the sole monomial is $1$. Their substituted images have $t$-exponents respectively $3b$, $3b+1$ and $3b+2$, so they are exactly the distinct monomialsThese are linearly independent. Consequently any degree-$m$ polynomial in $\ker \phi$, after reduction modulo $J$, has all its reduced coefficients zero. It belongs to $J$. Applying this to each homogeneous component givesThis also proves that the quotient embeds in the domain $k[s,t]$, so the displayed ideal is prime; no radicalization or saturation correction is concealed in the argument. Its image is the third Veronese subring ${⨁}_{m\ge 0}k[s,t{]}_{3m}$, and the calculation gives the Hilbert polynomial $3m+1$, consistent with degree three and genus zero.