= Solution
Write the target coordinates as $x_1,x_2,x_3,x_4$. Every image point satisfies $x_3=x_2^2$ and $x_4=x_1x_2$. Conversely any point satisfying these equations is obtained by taking $t=x_2$, $u=x_1$. Hence the image is the closed <affine variety>
$$
\boxed{X=V(x_3-x_2^2,\ x_4-x_1x_2)}.
$$
The substitution homomorphism $k[x_1,x_2,x_3,x_4]\to k[u,t]$ has kernel exactly the ideal $J=(x_3-x_2^2,x_4-x_1x_2)$: reduce any <polynomial> by these two relations to a <polynomial> in $x_1,x_2$, which vanishes after substitution only if it is zero. Equivalently the quotient by $J$ is $k[x_1,x_2]$, so $J$ is prime. Since $k$ is algebraically closed, or directly by the same polynomial-vanishing argument, \b[$I(X)=J$].
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