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Suppose $R$ is a domain and $I=aR$ be a non-zero principal ideal. Then, every element of $I$ has a unique representation, for if $ra=sa$ then $(r-s)a=0$. Since, $a\neq 0$ and $R$ is a domain, we have, $r-s=0$ and thus, $r=s$.

Can we extend this to non-pricipal ideals. That is, given an ideal $J=(a_1,...,a_n)R$ where $a_1,...,a_n$ are minimal generators of $J$, does every element of $J$ have a unique representation as a $R$ linear combination of $a_1,...,a_n$?

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up vote 2 down vote accepted

The answer is no: for example, take $R=\mathbb{C}[x,y]$, and let $I=(x,y)$. Then the element $xy\in I$ can be expressed as either $xy=y\cdot x+0\cdot y$ or as $xy=0\cdot x+x\cdot y$.

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Thanks for the quick response. Will the answer be true if every subset of the minimal generators generates a prime ideal? I am thinking we can somehow reduce to the principal case and still preserve the domain assumption. –  BMI Oct 10 '11 at 2:02
    
@BMI: The example I gave satisfies that condition: the set $\{x,y\}$ is a minimal set of generators for $I=(x,y)$, and $(x)$, $(y)$, and $(x,y)$ are all prime ideals in $\mathbb{C}[x,y]$. –  Zev Chonoles Oct 10 '11 at 2:08
    
Oh, sorry, my bad. –  BMI Oct 10 '11 at 2:52
    
+1 @BMI Additional Commentary: Zev's answer shows that $\{x,y\}$ is not a $\mathbb{C}[x,y]$-basis for the $\mathbb{C}[x,y]$-module $(x,y)$. You might wish to prove that, in fact, $(x,y)$ is not a free $\mathbb{C}[x,y]$-module. (Hint: tensor over $\mathbb{C}[x,y]$ with the fraction field $\mathbb{C}(x,y)$ of $\mathbb{C}[x,y]$.) Can you prove that $(x,y)$ is not even a flat $\mathbb{C}[x,y]$-module? If you know some algebraic geometry (e.g., Hilbert's nullstellensatz), then interpret these questions (and their answers) geometrically. –  Amitesh Datta Oct 10 '11 at 5:14
    
@BMI I think that I should perhaps expand on my hints. Firstly, in order to prove that $(x,y)$ is not a free $\mathbb{C}[x,y]$-module, can you answer the question: what is the dimension of $(x,y)\otimes_{\mathbb{C}[x,y]} \mathbb{C}(x,y)$ as a $\mathbb{C}(x,y)$-vector space? Secondly, in order to prove that $(x,y)$ is not even a flat $\mathbb{C}[x,y]$-module, do you remember the equational criterion for flatness (cf. the first theorem in chapter 2 of Hideyuki Matsumura's Commutative Algebra)? –  Amitesh Datta Oct 10 '11 at 5:24
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