I have the ideal $$\langle x^2, xy^3\rangle$$
where I've found a minimal primary decomposition to be $(x) \cap (x^2,y^3)$. Now I've been asked to find a second minimal primary decomposition but I'm not sure of how to go about it? I know that the associated primes of a second decomposition will have to be the same as the first ($(x), (x,y)$), but I'm not sure how to find this second minimal decomposition.
Notice that \begin{align} \langle x^2, xy^3 \rangle = \langle x^2, xy^3, xy^7 \rangle &= \langle x^2, xy^3, x\rangle \cap \langle x^2, xy^3,y^7\rangle \\ &= \langle x\rangle \cap \langle x^2, xy^3,y^7\rangle \end{align} Also the term $\langle x^2, xy^3,y^7\rangle$ is irreducible (check) in a noetherian ring and so is primary.
The trick appears to be adding another term that is 'un-needed' and splitting and simplifying.
Edit: The power of $7$ in $xy^7$ could be replaced by any integer $\geq 3$, I just like 7.