Affine algebraic subset of $\mathbb{A}_k^4$ How do I go about proving the subset $V = \{(s^3, s^2t, st^2, t^3)\mid s, t \in k\}$ is an affine algebraic subset of $\mathbb{A}_k^4$ and find $\mathbb{I}(V) \subset k[x_0, x_1, x_2, x_3]$?
 A: Consider the ideal $I = \langle x_0x_2 - x_1^2, x_0x_3 - x_1x_2, x_1x_3 - x_2^2 \rangle$. Denote $W = \mathbb{V}(I)$. Clearly $V \subset W$; the claim is $W \subset V$. Let $p = (a_0, \dots, a_3)$ be an element of $W$. If $a_0 = a_3 = 0$, then $a_1^2 = a_0a_2 = 0$ and $a_2^2 = a_1a_3 = 0$ so that $p = (0, 0, 0, 0)$, which is in $V$. Therefore assume $a_0 \neq 0$ or $a_3 \neq 0$; without loss of generality $a_0 \neq 0$. Denote by $s \in k$ any cube root of $a_0$ and denote $t = sa_1/a_0 = a_1/s^2$. Then $a_1 = s^2t$, $a_2 = (a_0a_2)/a_0 = a_1^2/a_0 = s^4t^2/s^3 = st^2$, and $a_3 = (a_0a_3)/a_0 = (a_1a_2)/a_0 = s^3t^3/s^3 = t^3$. So $p = (s^3, s^2t, st^2, t^3)$, which is in $V$. Thus $V = \mathbb{V}(I)$.
Every $I$-congruence class of elements in $k[x_0, x_1, x_2, x_3]$ contains an expression, $f = a(x_0, x_3) + b(x_0, x_3)x_1 + c(x_0, x_3)x_2$, for unique polynomials $a(x_0, x_3)$, $b(x_0, x_3)$, $c(x_0, x_3) \in k[x_0, x_3]$. Consider the $k$-algebra homomorphism$$\phi: k[x_0, x_1, x_3, x_3] \to k[s, t],$$$$x_0 \mapsto s^3, \text{ }x_1 \to s^2t,\text{ }x_2 \mapsto st^2,\text{ }x_3 \mapsto t^3.$$The image $\phi(f)$ is $a(s^3, t^3) + b(s^3, t^3)s^2t + c(s^3, t^3)st^2$. Gathering monomials whose $s$ and $t$ exponent are congruent modulo $3$, $\phi(f) = 0$ if and only if $a(s^3, t^3) = b(s^3, t^3) = c(s^3, t^3) = 0$, i.e. if and only if $f = 0$. So $\phi$ determines an injective $k$-algebra homomorphism $k[x_0, \dots, x_3]/I \to k[s, t]$. Since $k[s, t]$ is an integral domain, also $k[x_0, \dots, x_3]/I$ is an integral domain. Hence $I$ is a prime ideal. By the Strong Nullstellensatz, $\mathbb{I}(V) = \text{rad}(I) = I$.
A: KDong has already given the answer. There are some points that might help understand  how the equations describing $V$ were  arrived at. Note that your set $V$ is defined by points whose co-ordinates are all monomials of degree 4 in $s$ and $t$. They are ordered in ascending powers of $t$. 
In general one can do the same with sets defined by all monomials of fixed degree $d$.(They will define an affine algebraic subset in an affine space of dimension $d+1$). 
In this ordering take two  monomials $M_{i-1}$ and $M_{i+1}$.
Sum of their $s$-degrees (or $t$-degrees) is clearly twice that of $M_i$. So $M_i^2=M_{i-1}M_{i+1}$. This explains the equations of the kind  $x_i^2- x_{i-1}x_{i+1}=0$. Same way one can see $M_iM_j=M_kM_l$ if $i+j=k+l$. So this explains the equations of the kind $x_0x_3-x_1x_2=0$. The beauty is that these equations suffice. (Note that all the equations are quadratic). 
