The problem is to describe vanishing ideal $I \in K[x,y]$ of algebraic set V($y-x^2$), where $K$ could be arbitrary field, not necessary algebraically closed.
It's quite obvious in a case of algebraically closed field, because $<y-x^2>$ is prime and one may use Hilbert zeros theorem. What about non-algebracially closed situation ?
Right answer!
By definition $$ V(y-x^2)=\left\{(x,y)\in\mathbb{A}^2_{\mathbb{K}}\mid y-x^2=0\right\}, $$ or in other words $V(y-x^2)$ is the image of the map $$ \varphi:t\in\mathbb{A}^1_{\mathbb{K}}\to\left(t,t^2\right)\in\mathbb{A}^2_{\mathbb{K}}; $$ the relevant pull-back is the surjective homomorphism of rings: $$ \varphi^{*}:\mathbb{K}[x,y]\to\mathbb{K}[t]\mid\varphi^{*}(x)=t,\varphi^{*}(y)=t^2, $$ then: $$ I\left(V\left(y-x^2\right)\right)=\ker\varphi^{*}\supseteq\left(y-x^2\right). $$ Because $$ \dim_{Krull}\mathbb{K}[x,y]=2,\,\text{ht}\left(y-x^2\right)=1,\,\mathbb{K}[x,y]_{\displaystyle/\ker\varphi^{*}}\cong\mathbb{K}[t] $$ and $\mathbb{K}[t]$ is not a field, $\ker\varphi^{*}$ is not a maximal ideal and then $$ \ker\varphi^{*}=\left(y-x^2\right)! $$
Wrong answer!
A "hand proof" is the following!
If $(y-x^2)$ is not a prime ideal then $y-x^2$ is not an irreducible polynomial, and there exist \begin{gather*} a_1y+b_1x+c_1,a_2y+b_2x+c_2\in\mathbb{K}[x,y]\mid y-x^2=(a_1y+b_1x+c_1)(a_2y+b_2x+c_2)\\ y-x^2=a_1a_2y^2+(a_1b_2+a_2b_1)xy+b_1b_2x^2+(a_1c_2+a_2c_1)y+(b_1c_2+b_2c_1)x+c_1c_2\iff\\ \iff\begin{cases} a_1a_2=0\\ a_1b_2+a_2b_1=0\\ b_1b_2=1\\ a_1c_2+a_2c_1=1\\ b_1c_2+b_2c_1=0\\ c_1c_2=0 \end{cases} \end{gather*} and easily one finds a contradiction. Q.E.D. $\Box$