How to find the primary decomposition of this ideal? How does this fit within the algebro-geometric picture? I'm new to classical algebraic geometry and I'm trying to play around with its objects and making examples before diving into the theory itself, just trying to see what is going on.
For instance, consider the algebraic variety defined over $\mathbb{R}$ by the equation
\begin{equation}
f(x,y)=y^{3}+2x^{2}y-4y-x=0.
\end{equation}
From what I understood so far,  we investigate 'sub-objects' of this algebraic variety by looking at ideals in $\mathbb{R}[x,y]$ containing the principal ideal $I=(y^{3}+2x^{2}y-4y-x)$. For example, the maximal ideals of $\mathbb{R}[x,y]$ containing $I$ correspond to the sub-objects points of the variety.
More generally, if $J=(f_{1}(x,y),\dots,f_{n}(x,y))$ is an ideal containing $I$, then an solution for the system of equations $f_{1}(x,y)=\dots f_{n}(x,y)=0$ gives a solution for $f(x,y)=0$, and hence this ideal defines some subset of points of the original variety.
So, since maximal ideals correspond to points, the natural thing is to wonder what happens inside the algebraic variety for more general prime ideals or even for some other kind of ideals.
Is this correct so far? Recall that I'm trying to make first a picture of what algebraic geometry does.
Now, from commutative algebra we have the result that in a Noetherian ring, one always has a minimal primary decomposition for an ideal. I want to find the primary decomposition for this specific $I=(y^{3}+2x^{2}y-4y-x)$ to see how this fits within the geometrical picture.
How can I do that? Is there some algorithmical procedure for finding such a decomposition?
 A: In general:in algebraic geometry $k[X]=k[x_1,...,x_n]$, $$V(I)=\{(a_1,...,a_n)\in k^n; f(a_1,..,a_n)=0 \forall f\in k[X] \}$$
$$I(X)=\{f\in k[X]; f(x_1,...,x_n)=0 \forall (x_1,...,x_n)\in X \}$$
Let be $I,J$ ideals in $ k[X]$ and $I \subseteq J$, then $V(J) \subseteq V(I)$ . In question we have $I\subseteq J=(x,y)$, then $V(J)=(0,0)$ is a solution of $f(x,y)=y^3+2x^2y-4y-x$.
If $k$ ia an algebraical closed, then by Hilbert Nullstellensatz theorem we have  bijections from maximal ideals of $k[X]$ and points of $\mathbb A^n_k $ , prime ideals of $k[X]$ and irreducible subvarieties of $\mathbb A^n_k$ ,radical ideals of $k[X]$ and  subvarieties of $\mathbb A^n_k$ .
but if $k$ isnot algebraical closed, for example $\mathbb R$, and $I=(x^2+1)$ is maximal ideal in $\mathbb R[x]$, then $V(I)= \phi$. 
In general $J\subseteq I(V(J))=I(X)$ and if $X=X_1 \cup X_2$ , then $I(X_1\cup X_2)=I(X_1) \cap I(X_2)$
A: You are asking many questions. I will try to give short answers:
For example, the maximal ideals of ℝ[x,y] containing I correspond to the sub-objects points of the variety: This is wrong as it depends on the Nullstellensatz which only holds over algebraically closed fields.
Is this correct so far? Recall that I'm trying to make first a picture of what algebraic geometry does.: How about a plot of the curve?
I want to find the primary decomposition for this specific $I=(y^3+2x^2y−4y−x)$ to see how this fits within the geometrical picture. The curve is irreducible as an algebraic variety since your polynomial generates a prime ideal. Therefore primary decomposition cannot tell you more about the curve.  However, primary decompositions can be computed, for example with Macaulay2 or Singular.
Finally, whenever you are interested in questions over the real numbers, you should look into real algebraic geometry, which is quite different from classical algebraic geometry.
