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Let $F$ be a field and $R=F[X,Y]$ be the polynomial ring of two variables over $F$. Let $I\subset R$ be the ideal generated by $X^2$ and $XY$, find the associated prime ideals of $R/I$.

I'm really stuck on this one, I'm trying to show that if $P\in Ass(R/I)$ then there is an injective map $i:R/P\rightarrow R/I$ and I'm trying to find such a map. I'm not sure if I'm on the write track or not. Any ideas. Thank you!

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Hint: There are exactly two such primes! –  neilme Feb 13 '13 at 3:12
    
@neilme Can you tell me how do you know that? –  i.a.m Feb 13 '13 at 3:13
    
@neilme, I know that if $P$ is prime then $Ass(R/P)=\{P\}$ are you telling me that $Ass(R/<X^2,XY>)=\{<X^2>,<XY>\}$? –  i.a.m Feb 13 '13 at 3:44
    
No; neither of the ideals you listed in the comment are prime. Use Jim's outline below; it will get you to where you need to go. –  neilme Feb 13 '13 at 13:07

2 Answers 2

Hint: $(x^2, xy) = (x) \cap (x, y)^2$

Read up on primary decompositions.

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Thank you for the hint, but I haven't learned about primary decompostoin yet. is there any other way. –  i.a.m Feb 13 '13 at 3:47
    
To start, what are all the prime ideals in the ring $R/I$? You should be able to list all of them. –  Jim Feb 13 '13 at 3:54
    
I'm not sure how to do it –  i.a.m Feb 13 '13 at 3:55
    
Do you know the correspondence between prime ideals of $R$ and prime ideals of $R/I$? –  Jim Feb 13 '13 at 3:56
    
Yes, prime ideals of R/I are prime ideals of R containing I, right? –  i.a.m Feb 13 '13 at 3:57

As suggested, the key here is the Primary Decomposition Theorem. Here we cite it for clarity:

$\mathit{Theorem}:$ Let $R$ be a Noetherian ring. Then every proper ideal $I$ in $R$ has a minimal primary decomposition. If $I=\bigcap^{m}_{i=1}Q_{i}=\bigcap^n_{i=1}Q^{\prime}_{i}$ are two minimal primary decompositions for $I$ then the sets of associated primes in the two decompositions are the same: $\left\{\mathrm{rad}\,Q_{1},...,\mathrm{rad}\,Q_{m}\right\}=\left\{\mathrm{rad}\,Q^{\prime}_{1},...,\mathrm{rad}\,Q^{\prime}_{n}\right\}$. Moreover, the primary components $Q_{i}$ belonging to the minimal elements in this set of associated primes are uniquely determined by $I$.

Now let $I=(x^{2},xy)$ in $R[x,y]$. Then $(x^{2},xy)=(x)\cap (x,y)^{2}=(x) \cap (x^{2},y)$ are two minimal primary decompositions for $I$. The associated primes of $I$ are then $(x)$ and $\mathrm{rad}\,((x,y)^{2})=\mathrm{rad}\,((x^{2},y))=(x,y)$. We note that $(x)$ is an isolated prime since $(x) \subset (x,y)$, and $(x,y)$ is an embedded prime.

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