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I have a question about the definition of a seed of a cluster algebra. It is said that a seed is a pair $(R, u)$, where $R$ is a quiver with $n$ vertices, $u = \{u_1, \ldots, u_n\}$ is a free generating set of the field $Q(x_1, \ldots, x_n)$, see Page 10 of the paper.

I think here $u_i$ is in terms of $x_1, \ldots, x_n$ and $u_1, \ldots, u_n$ generate $Q(x_1, \ldots, x_n)$ freely. Is this true? Thank you very much.

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I do not understand what you are asking... The $u_i$ are elements of $Q(x_1,\dots,x_n)$ so they are automatically «in terms of $x_1$, $\dots$, $x_n$». Since the set $u$ is to be, as you write, a free generating set, in particular it generates. What is it that you want to know if it is true? – Mariano Suárez-Alvarez Aug 14 '12 at 17:45
@Mariano, yes. For example, I think $u_1$ can be $x_1x_2+x_3$. Every element of $Q(x_1, \ldots, x_n)$ can be written as a sum of some products of $u_1, \ldots, u_n$. Is this true? – LJR Aug 14 '12 at 18:10
Yes what? :-) I do not understand what you are asking! But no, not every element of $Q(x_1,\dots,x_n)$ can be written as a sum of some products of $u_1$, $\dots$, $u_n$: for example, if $u_i=x_i$ this is clearly not true. – Mariano Suárez-Alvarez Aug 14 '12 at 18:13
@Mariano, $x_i$ is a sum of products of $u_i=x_i$ (only one factor and one summand). – LJR Aug 14 '12 at 18:53
You have not yet make clear what it is you are trying to ask. I suggest you edit your question so that it ends with «What I want to to know is if SOMETHING». – Mariano Suárez-Alvarez Aug 15 '12 at 0:50
up vote 1 down vote accepted

Each $u_i$ is a rational function in the $x_i$, with coefficients in $\mathbb{Q}$ - that's what it means to be an element of $\mathbb{Q}(x_1,\dotsc,x_n)$.

They set of $u_i$ also freely generates $\mathbb{Q}(x_1,\dotsc,x_n)$ as a field extension of $\mathbb{Q}$; another way of saying this is that every element of $\mathbb{Q}(x_1,\dotsc,x_n)$ can be written uniquely as a rational function in the $u_i$.

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No, to say that the $u_i$ form a transcendence basis of the field over $\mathbb Q$ is weaker than saying that they generate this field. A transcendence base for a field $F$ might generate a subfield of $F$ over which $F$ is algebraic. (Your second "perhaps the most useful" reformulation is correct.) – Andreas Blass Aug 24 '13 at 19:26
@AndreasBlass Corrected - I usually have $\mathbb{C}$ in place of $\mathbb{Q}$, which is where that came from. – Matthew Pressland Aug 29 '13 at 8:19

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