Ideal for two polynomials in three variables Consider the set $B=\{(t^2,t^3,t^4)\mid t\in \mathbb{C}\}$. It is a subvariety of $\mathbb{C}^3$, because it is equal to $V(y^2-x^3,z-x^2)$.
How can we find the ideal $I(B)$? I think it is $I(\langle y^2-x^3,z-x^2\rangle)$. It is clear that $I(\langle y^2-x^3,z-x^2\rangle)\subseteq I(B)$, because any polynomial $a(x,y,z)(y^2-x^3)+b(x,y,z)(z-x^2)$ vanishes on $B$, but how can we prove the other inclusion, namely $I(B)\subseteq I(\langle y^2-x^3,z-x^2\rangle)$?
 A: Let $I=\langle Y^2-X^3,Z-X^2\rangle\subset \mathbb C[X,Y,Z]$ .
In the quotient $\frac {\mathbb C[X,Y,Z]}{I}=\mathbb C[x,y,z]$, where $z=x^2$ and $y^2=x^3$, every element can be written $f(x,y,z)=a(x)+b(x)y$ where $a(X), b(X)\in \mathbb C[X]$ are polynomials: just replace in $f(x,y,z)$ every occurrence of $z$ by $x^2$,  any occurrence of $y^2$ by $x^3$ and repeat.
Hence every polynomial $f(X,Y,Z)\in \mathbb C[X,Y,Z]$ can be written as $$f(X,Y,Z)=a(X)+b(X)Y+i(X,Y,Z) \;\text {with}\; i(X,Y,Z)\in I $$ So, to say that $f(X,Y,Z)$ vanishes on $B$ means that $f(t^2,t^3,t^4)=a(t^2)+b(t^2)t^3+i(t^2,t^3,t^4)=a(t^2)+b(t^2)t^3$  must be zero for all $t\in \mathbb C$ (recall that since $i$ is in $I$, $i$  vanishes on $B$).     
And now for the coup de grâce:    
All the monomials in $a(t^2)$ have even exponents whereas all the monomials in $b(t^2)t^3$ have odd exponent : so the identity $a(t^2)+b(t^2)t^3\equiv 0$ can only hold if $a(X)=b(X)=0 \in \mathbb C[X]$.
But then $f(X,Y,Z)=a(X)+b(X)Y+i(X,Y,Z)=i(X,Y,Z)$ belongs to $I$:
we have proved that any polynomial $f(X,Y,Z)$ vanishing on $B$ satisfies $f(X,Y,Z)\in I$ and thus that $I:=\langle Y^2-X^3,Z-X^2\rangle=I(B)$.
A: On $\mathbb N_0^3$ define the following order:
$$\tag1(i,j,k)<(i',j',k'):\Leftrightarrow ( k<k'\lor(k=k'\land(j<j'\lor(j=j'\land i<i')))).$$
This is a well-order on $\mathbb N_0^3$.
For $0\ne f(x,y,z)=\sum_{i,j,k}a_{i,j,k}x^iy^jz^k\in\mathbb C[x,y,z]$ define
$$ v(f)=\max\{\,(i,j,k)\mid a_{i,j,k}\ne 0\,\}$$
where the maximum is taken with respect to the order given by $(1)$ (and exists because at least one and only finitely many coefficients are nonzero).
Assume $I(B)\setminus\langle y^2-x^3,z-x^2\rangle\ne \emptyset$ and let $f(x,y,z)\in I(B)\setminus\langle y^2-x^3,z-x^2\rangle$ minimize $v$.
If $(i,j,k)=v(f)$ then $k\ge 1$ would imply that 
$$f_1(x,y,z):=f(x,y,z)-a_{i,j,k}x^iy^jz^k+a_{i,j,k}x^{i+2}y^jz^{k-1}\in I(B)\setminus\langle y^2-x^3,z-x^2\rangle$$
with $v(f_1)<(i,j,k)$, contradiction. Therefore $k=0$.
Next, $j\ge 2$ would imply that 
$$f_2(x,y,z):=f(x,y,z)-a_{i,j,0}x^iy^jz^k+a_{i,j,0}x^{i+3}y^{j-2}z^k\in I(B)\setminus\langle y^2-x^3,z-x^2\rangle$$
with $v(f_2)<(i,j,k)$, contradiction. Therefore $j\le 1$.
Hence we can write 
$$f(x,y,z)= g(x)+h(x)y$$
with $g,h\in\mathbb C[x]$.
For all $t\in\mathbb C$ we have $f(t^2,t^3,t^4)=g(t^2)+h(t^2)t^3=0$. 
Hence the polynomial $g(t^2)+h(t^2)t^3\in\mathbb C[t]$ is the zero polynommial.  As $g(t^2)$ has only even powers and $h(t^2)t^3$ only odd powers, no cancelling occurs between the summands, i.e. $g$ and $h$ must each be the zero polynomial. But then $f$ is the zero polynomial, contrary to assumption.
