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I am trying to prove in an elementary way that $\mathbb{Q}$ is not a finitely presented $\mathbb{Q}[x_1,x_2,\dots]$-module.

According to this post, the general case for any non noetherian ring can be proven by showing that $\mathbb{Q}$ being finitely presented implies that $(x_1,x_2,\dots)$ would be a finitely generated ideal in $\mathbb{Q}[x_1,x_2,\dots]$, which is clearly false. However the proof uses Schanuel's lemma which I am not familiar with.

Therefore I am wondering whether there is a more elementary proof to show the result for the $\mathbb{Q}[x_1,x_2,\dots]$-module $\mathbb{Q}$?

EinStone
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    If $\mathbb{Q}$ was finitely presented as$\mathbb{Q}[x_1,\dots!]$-module, then we have $\varphi:\mathbb{Q}\stackrel{\cong}{\rightarrow}\mathbb{Q}[x_1, \dots]^n/I$ where $I$ is a finitely generated $\mathbb{Q}[x_1,\dots]$-module. However, if $I$ is finitely generated, then the generators only contain finitely many variables. Thus, there exists $x_n$ which does not appear in the generators. We have that $\varphi$ is a bijection and $\mathbb{Q}[x_1,\dots]$-linear and thus also $\mathbb{Q}$-linear. Hence, $\varphi$ is also an isomorphism as $\mathbb{Q}$-vector spaces. – Severin Schraven Dec 31 '20 at 17:15
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    Now we have the subspace generated by $(x_n, 0, \dots, 0)$ and $(1, 0,\dots,0)$ which are linearly independent over $\mathbb{Q}$ and thus we get a contradiction as $\mathbb{Q}$ would have dimension at least $2$ over the rationals. – Severin Schraven Dec 31 '20 at 17:19
  • Thanks for the argument. Could it not be that $x_n - 1 \in I$ and thus $(x_n, 0, \dots, 0)=(1, 0, \dots, 0)$ in $I$? – EinStone Dec 31 '20 at 17:39
  • Assume $x_n -1\in I$. Then there exists $f_1, \dots, f_n, g_1,\dots, g_n\in \mathbb{Q}[x_1, \dots]$ where $g_1,\dots, g_n$ do not depend on $x_n$ and such that $$x_n-1=\sum_{j=1}^n f_j g_j.$$ – Severin Schraven Dec 31 '20 at 18:18
  • I am sorry, I need to think how I could show the contradiction. Maybe we also should replace $1$ by $x_n^2$. – Severin Schraven Dec 31 '20 at 18:24
  • I suppose you assume that $g_1,\dots,g_n \in I$. How do you know that $g_i$ does not depend on $x_n$? I am thinking for example that $I$ could be generated by $(x_n -1, x_2,x_5)$, then $x_n$ is not a generator of $I$, but $x_n -1 \in I$ – EinStone Dec 31 '20 at 18:24
  • No, I am not just assuming that $x_n$ is not a generator, none of the generators depend on $x_n$. More rigorously, the generators are out of $\mathbb{Q}[x_1,\dots, x_{n-1}]$. – Severin Schraven Dec 31 '20 at 18:27
  • Ok if we set $x_n=0$, then we get $$-1=\sum_{j=1}^m \tilde{f}_j g_j$$ where $\tilde{f}_j$ equal to $f_j$ evaluated at $x_n=0$. However, then $I=(g_1, \dots, g_m)=\mathbb{Q}[x_1,\dots]$. – Severin Schraven Dec 31 '20 at 18:37
  • ok makes sense, thank you! – EinStone Dec 31 '20 at 18:38
  • I'll write it down rigorously at some point (now everything is handwaving, e.g. $I$ should not be an ideal but a submodul). Or you could try to do it yourself. – Severin Schraven Dec 31 '20 at 18:40
  • I will write a solution as an answer, could you proof read it? – EinStone Dec 31 '20 at 18:42
  • Sure, I'll have a look. I am curious whether my idea will work out in the end. – Severin Schraven Dec 31 '20 at 18:43

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Assume that $\mathbb{Q}$ is finitely presented. Then there exists $m \in \mathbb{N}$ and a finitely generated submodule $K \subset \mathbb{Q}[x_1,x_2,\dots]^m$, s.t. $\mathbb{Q} \cong \mathbb{Q}[x_1,x_2,\dots]^m /K$. Let $\varphi: \mathbb{Q} \to \mathbb{Q}[x_1,x_2,\dots]^m /K$ be such an isomorphism.

$\varphi$ is bijective and $\mathbb{Q}[x_1,x_2,\dots]$-linear, thus it is also $\mathbb{Q}$-linear, hence a $\mathbb{Q}$-linear vector space isomorphism. Thus $\dim_\mathbb{Q}(\mathbb{Q})=\dim_\mathbb{Q}(\mathbb{Q}[x_1,x_2,\dots]^m /K)$.

Since $K$ is finitely generated there exist $i_1,\dots,i_n \in \mathbb{N}$ and $f_1,\dots,f_s \in\mathbb{Q}[x_{i_1},\dots,x_{i_n}]^m$ with $K= \langle f_1,\dots,f_s\rangle$. Choose $x_j \notin \{x_{i_1},\dots,x_{i_n}\}$. Then $[(x_j,0,\dots,0)]$ and $[(x_j^2,0,\dots,0)]$ are $\mathbb{Q}$-linearly independent elements in $\mathbb{Q}[x_1,x_2,\dots]^m /K$ and so $\dim_\mathbb{Q}(\mathbb{Q}[x_1,x_2,\dots]^m /K) \geq 2$, which contradicts $\dim_\mathbb{Q}(\mathbb{Q})=1$.

EinStone
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    You cannot say that $K$ is a subspace of $\mathbb{Q}[x_{i_1}, \dots x_{i_n}]$, you can only say that this is true for the generators. Then you still need to show why the two vectors are independent over the rationals. – Severin Schraven Dec 31 '20 at 18:59
  • Thanks, I edited the first part in and will see to write up linear independence later. – EinStone Dec 31 '20 at 19:14
  • Like the OP, you are making some assumptions about how $\Bbb{Q}$ is to be understood as a module over $\Bbb{Q}[x_1, x_2, \ldots]$. The intention is probably that the action of the ring $\Bbb{Q}[x_1, x_2, \ldots]$ on $\Bbb{Q}$ is one of the many actions that extends the natural action of $\Bbb{Q}$ on $\Bbb{Q}$, in which case I agree with your answer. – Rob Arthan Dec 31 '20 at 22:19
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    Indeed, I (OP) was assuming the "standard" action $f(x)\cdot a = f(0)a$. – EinStone Jan 01 '21 at 10:53