Can anyone give me an idea about the primary decomposition of the ideal $I=(x^3y,xy^4)$ of the ring $R=k[x,y]$?

I am trying to connect the primary decomposition with the set Ass(R/I) which i have find before..


1 Answer 1


If $I$ is a monomial ideal with a generator $ab$ (where $a$ and $b$ are coprime), say $I = (ab) + J$, then $I = ((a) + J) \cap ((b) + J)$. Applying this recursively gives a primary decomposition for any monomial ideal:

\begin{align} I &= (x^3y, xy^4) = (x^3, xy^4) \cap (y, xy^4) \\ &= (x^3, x) \cap (x^3, y^4) \cap (y, x) \cap (y, y^4) \\ &= (x) \cap (x^3, y^4) \cap (x, y) \cap (y) \\ &= (x) \cap (x^3, y^4) \cap (y) \end{align}

  • $\begingroup$ i cant understand why the (x,y) is an associated prime...i cant find an element of k[x,y]/I , which ann(m)=(x,y)...could you tell me your opinion? $\endgroup$ Jan 6, 2014 at 0:03
  • $\begingroup$ @georgeson1960: have you tried $x^2y^3$? $\endgroup$
    – Youngsu
    Jan 6, 2014 at 0:12
  • $\begingroup$ you mean that ann(x^2y^3 + I)=(x,y)...Sorry but i cant understand that,could you explain to me Younsu? $\endgroup$ Jan 6, 2014 at 0:19
  • 1
    $\begingroup$ @georgeson1960: Youngsu is right, the image of $(x,y)$ in $k[x,y]/I$ is the annihilator of $x^2y^3$ - just check that $x \cdot x^2y^3, y \cdot x^2y^3 \in I$, and note that $(x,y)$ is maximal $\endgroup$
    – zcn
    Jan 6, 2014 at 2:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .