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The question is self-explanatory. If we have a an R-module $V$, where $R$ is a ring, every element in $V$ can be represented as $r*x$ [by division algorithm], where $r$ is any element from $R$ and $x$ is the smallest element in the module. So is the whole of the module just $(x)$?

Thanks in advance

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    What do you mean by "smallest element in the module"? –  Mar 28 '13 at 05:35
  • In an ordered R-module $V$, let $X$ be the smallest element. Let us divide any element of the module $V$ with $X$. $v=r_1.*x+r$, where $r$ is the remainder, and expected to lie between $0$ and $x$. But the module can't have any smaller positive element than $X$. Hence, $r=0$, and any element of the module can be expressed as a scalar multiple of $x$. –  Mar 28 '13 at 05:51
  • $v$ is a member of the R-module $V$, and $r_1$ is an element of the ring $R$. –  Mar 28 '13 at 05:54
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    Hoping to make the penny drop. Let $R=\mathbb{Z}[x]$ and let $V$ be the submodule of $R$ consisting of the polynomials that have an even value at $x=0$. We have $2\in V$ and $x\in V$. It is easy to see that $2$ and $x$ generate all of $V$ as an $R$-module. Which is smaller $2$ or $x$? Remember that $x$ has no "value", it is just an indeterminate. – Jyrki Lahtonen Mar 28 '13 at 06:03
  • I had thought the division algorithm pre-supposes that only polynomials divde polynomials and integers, integers. You may say $2x^2+3=3*1+2x^2$, where 3 is the divisor and $2x^2$ is the remainder, but I don't think that'll culminate in any fruitful discussion. So you're right. We can't decide whether $x$ is smaller or $2$. But that's not the point. –  Mar 28 '13 at 06:21
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    But your argument relies on your ability to magically compare elements of any module - with a view of finding the smallest. So I repeat, what is the smallest element of that module? – Jyrki Lahtonen Mar 28 '13 at 19:29

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This is certain false. A module which is singly-generated, it is called cyclic. A perfect example of a non-cyclic module is $R^2$ (assuming $R$ is commutative, or more generally has the IBN property).

Alex Youcis
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  • Where am I going wrong? Let $ v_1,v_2 \cdots, v_n $ be the elements of the module $V$, over the ring $R$. Let the smallest element of $V$ be $x$. $ v_k=r*x+y $, where $y$ is the remainder. $y$ is obviously in the module, and hence, all elements of $V$ can be expressed as multiples of $y$. –  Mar 28 '13 at 05:46
  • @AyushKhaitan: Your argument assumes there is an ordering on $V$. Why are you assuming that every $R$-module has a compatible ordering? – Zev Chonoles Mar 28 '13 at 05:56
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    @Ayush: what is the "smallest element of $V$"? – Qiaochu Yuan Mar 28 '13 at 05:58
  • @ZevChonoles: If we assume that there is ordering in V, which I understand to not be the general case, am I correct? –  Mar 28 '13 at 06:06
  • @QiaochuYuan: I just answered this question in the comments section of the main question. You may refer to it. –  Mar 28 '13 at 06:07