I'm anaysing the primary decomposition of $I=(x^3(y-1), x^2y(y-1), xy^2(y-1))\subset k[x,y]$ (where $k$ is an algebraicaly closed field). Here is what I've found:
$$I=(x)\cap(y-1)\cap(x, y)^3$$
If this is correct, it means $(x), (y-1)$ and $(x, y)$ are the associated primes of $I$, where $(x), (y-1)$ are isolated and $(x,y)$ is embedded.
Here's a remark from page $52$ in Atiyah-Macdonald's Introduction to Commutative Algebra:
The names isolated and embedded come from geometry. Thus if $A=k[x_1,...,x_n]$, where $k$ is a field, the ideal $\mathfrak a$ gives rise to a variety $X \subset k^n$. The minimal primes $p_i$ corresponds to the irreducible components of $X$ and the embedded primes corresponds to subvarieties of these, i.e., varieties embedded in the irreducible components.
By looking at $V(I)$ (which is the union of the lines $x=0$ and $y=1$), we get that $V(x)=\{x=0\}$ and $V(y-1)=\{y=1\}$ are the irreducible components (which is what I expected), but the embedded variety is $V(x, y)=\{(0,0)\}$, which was confusing to me. By his explanation, I was expecting to find the embedded subvariety $V(x,y-1)=\{(0,1)\}$, which is embedded in both components, so this point $(0,0)$ seemed kind of random to me.
Why is $(0,0)$ special, geometricaly speaking? Why does it show up instead of $(0,1)$?
You ask: "Why is $(0,0)$ special?" - it is not special, your ideal is just so that the corresponding scheme simply has an embedded component at $(0,0)$. For example, if you consider the ideal $$J:=(x)\cap(y-1)\cap(x,y-1)^2 = (xy^2-xy, x^2y)$$ instead, you get a different ideal which also has the components $\{y=1\}$ and $\{x=0\}$ as well as an embedded component at $(0,1)$. More generally, if you consider the ideal $$(x)\cap(y-1)\cap(x,y-b)^2,$$ you will have the same components $\{y=1\}$ and $\{x=0\}$ and an embedded component at $(0,b)$. You can have embedded component(s) anywhere on the lines $\{y=1\}$ and $\{x=0\}$ by simply constructing the corresponding ideal to be the appropriate intersection.