Elementary proof that $\mathbb{Q}$ is not a finitely presented $\mathbb{Q}[x_1,x_2,\dots]$-module 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 $\mathbb{Q}$?
 A: 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 $\dim_\mathbb{Q}(\mathbb{Q})=1$.
