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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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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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    @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*)?2011-10-10