# Show that $\mathbb{Q}$ is finitely generated $\mathbb{Q}[x_1,x_2,\dots]$-module

Let $$R=\mathbb{Q}[x_1,x_2,\dots]$$ and $$M=\mathbb{Q}$$. Show that $$M$$ is finitely generated $$R$$-module.

First of all, I don't understand how is $$M$$ an $$R$$-module; Indeed $$f(x_1,x_2,\dots)=x_1\in R, 1\in M$$ but $$1\cdot f(x_1,x_2,\dots)=x_1\notin \mathbb{Q}=M$$.

Is this a mistake in the question.

Assuming it's a mistake we need to change $$R\leftrightarrow M$$. I still don't know how to solve it because we need to find a finite basis $$\{m_1,\dots ,m_n\}\subseteq M$$ s.t. $$M=\{\sum_{i=1}^nr_im_i:r_i\in R\}$$.

The netural guess is to detemine $$m_i:=x_i$$ but this way $$x_{n+1}\notin \operatorname{span} \{x_1,\dots,x_n\}$$.

• Turning it around, the polynomial ring over a field is not finitely generated over this field. – Wuestenfux Aug 29 at 12:00

An $$R$$-module $$M$$ is an abelian group equipped with a ring homomorphism $$\varphi:\ R\ \longrightarrow\ \operatorname{End}(M).$$ In this particular case, a ring homomorphism $$\varphi:\ \Bbb{Q}[x_1,x_2,\ldots]\ \longrightarrow\ \operatorname{End}(\Bbb{Q}),$$ is uniquely determined by where the indeterminates $$x_i$$ are mapped, and an endomorphism of $$\Bbb{Q}$$ (as an abelian group!) is uniquely determined by where $$1\in\Bbb{Q}$$ is mapped. So $$\varphi$$ is uniquely determined by the values $$(\varphi(x_i))(1)\in\Bbb{Q}$$.
Note that here $$\varphi(x_i)$$ is an endomorphism of $$\Bbb{Q}$$ for each $$i$$. In your notation $$(\varphi(x_i))(1)=1\cdot x_i$$. Recall that the '$$\cdot$$' in $$1\cdot f(x_i)$$ stands for the 'multiplication' defined by the $$R$$-module structure on $$M$$; it is not the regular multiplication $$1\cdot f(x_i)\in\Bbb{Q}[x_1,x_2,\ldots]$$ because we are not in $$\Bbb{Q}[x_1,x_2,\ldots]$$, we are in $$\Bbb{Q}$$ with a $$\Bbb{Q}[x_1,x_2,\ldots]$$-module structure.
• I understand that there's an equivalent defenition for an $R$-module. But still, how could $\mathbb{Q}$ be a $\mathbb{Q}[x_1,x_2,...]$-module if $1\in \mathbb{Q}, x_1\in \mathbb{Q}[x_1,...]$ but $1\cdot x_1=x_1\notin \mathbb{Q}$? – J. Doe Aug 29 at 12:18
• The dot in the multiplication $1\cdot x_1$ is not multiplication in $\Bbb{Q}[x_1,x_2,\ldots]$. It denotes the module action of $\Bbb{Q}[x_1,x_2,\ldots]$ on $\Bbb{Q}$. When getting familiar with modules, it can be very illustrative to make this difference explicit by not writing $1\cdot x_1$, but in stead writing $\varphi_{x_i}(1)$, where $\varphi$ is the module homomorphism described above, and $\varphi_{x_i}:=\varphi(x_i)\in\operatorname{End}(\Bbb{Q})$. – Inactive - avoiding CoC Aug 29 at 12:21
• As a simpler example, the ring homomorphism $$\varphi:\ \Bbb{Q}[x]\ \longrightarrow\ \operatorname{End}(\Bbb{Q}):\ x\ \longmapsto\ (q\ \mapsto\ 2q),$$ defines a $\Bbb{Q}[x]$-module structure on $\Bbb{Q}$. With this structure you have $q\cdot x=\varphi_x(q)=2q$, and so in particular $1\cdot x=2$. – Inactive - avoiding CoC Aug 29 at 12:24
• To complete the answer, let $T:=\{\sum_{i=1}^n(\varphi(r_i))(m_i):r_i\in R\}$. We want to find $m_1,\dots,m_n\in\mathbb{Q}$ s.t. $\mathbb{Q}=T$. Now, we know that $\operatorname{id}_{\mathbb{Q}},0_{\operatorname{End}(M)}\in \operatorname{Im}(\varphi)$. Hence, if we choose for example $n=1,m_1=1$ then we have $0,1\in T$. But how can we choose $n,m_1,\dots,m_n$ s.t. $\forall q\in\mathbb{Q}, q\in T$? – J. Doe Aug 29 at 12:55
• You already note that $\operatorname{id}_{\Bbb{Q}}\in\operatorname{Im}(\varphi)$; indeed necessarily $\operatorname{id}_{\Bbb{Q}}=\varphi(1)$, for $1\in\Bbb{Q}[x_1,x_2,\ldots]$. Because $\varphi$ is a ring homomorphism it follows that $\varphi(q)=q\varphi(1)=q\cdot\operatorname{id}_{\Bbb{Q}}$ for all $q\in\Bbb{Q}\subset\Bbb{Q}[x_1,x_2,\ldots]$, and hence $$q=q\cdot\operatorname{id}_{\Bbb{Q}}(1)=(\varphi(q))(1).$$ So not only is $\Bbb{Q}$ finitely generated over $\Bbb{Q}[x_1,x_2,\ldots]$, it is cyclic because it is generated by $1\in\Bbb{Q}$. – Inactive - avoiding CoC Aug 29 at 13:13