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Show that $\Bbb Q$ is not a finitely generated $\Bbb Z$-algebra.

I know that $\Bbb Q$ is not a finitely generated $\Bbb Z$-module. From here how can I conclude that $\Bbb Q$ is not a finitely generated $\Bbb Z$-algebra?

Please help me in this regard.

Thank you in advance.

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    $\begingroup$ The largest prime dividing denominators doesn't increase by addition multiplication, or multiplication by an integer. $\endgroup$ – user553213 Apr 28 '18 at 20:49
  • $\begingroup$ Let's understand the fact in a simpler way. Suppose that the generating set for $\Bbb Q$ is a two elements set containing $\frac {1} {5}$ and $\frac {2} {3}$. Then what will happen? $\endgroup$ – Arnab Chatterjee. Apr 28 '18 at 21:05
  • $\begingroup$ My previous comment will happen. $\endgroup$ – user553213 Apr 28 '18 at 21:08
  • $\begingroup$ Would you please be more explicit? $\endgroup$ – Arnab Chatterjee. Apr 28 '18 at 21:09
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Let $\varphi\colon\mathbb{Z}[x_1,\dots,x_n]\to\mathbb{Q}$ be a ring homomorphism. I claim that $\varphi$ is not surjective.

Let $\varphi(x_k)=a_k/b_k$, for $k=1,\dots,n$. Choose a prime $p$ that does not divide $b_k$, for each $k$. Suppose that $\varphi(f)=1/p$. Denote by $\deg_k(f)$ the degree of $f$ as a polynomial in $x_k$ and let $$ m=\max_k\deg_k(f) $$ Then we can write $$ f(a_1/b_1,\dots,a_n/b_n)=\frac{q}{b_1^m\dots b_n^m} $$ where $q$ is an integer (prove it). Then from $\varphi(f)=1/p$ we derive $$ pq=b_1^m\dots b_n^m $$ which is a contradiction.

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Suppose $\mathbb Q$ is finitely generated as a $\mathbb Z$-module (in this case, $\mathbb Z$-algebra). By the Structure Theorem for modules (in this case, it is the theorem on finitely-generated abelian groups) and the fact that $\mathbb Q$ is torsion free, $\mathbb Q\cong \mathbb Z^r$ for some $r\geq 0$. Let $\varphi:\mathbb Q\rightarrow \mathbb Z^r$ be an isomorphism. Since $\varphi$ is a $\mathbb Z$-module homomorphism, if $n\in \mathbb Z$ is nonzero, then $\varphi(1)=\varphi\left(n\cdot \frac{1}{n}\right)=n\varphi\left(\frac{1}{n}\right)$. Let $\varphi(1)=(a_1,\ldots,a_r)$. Take $n=\mbox{gcd}(a_1,\ldots,a_r)+1$. Then $\varphi\left(\frac{1}{n}\right)\notin\mathbb Z^r$, a contradiction.

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Hint:

If we had $\;Q=\mathbf Z\biggl[\dfrac{m_1}{n_1},\dots,\dfrac{m_k}{n_k} \biggr]$, we could as well write $$Q=\mathbf Z\biggl[\frac1N \biggr], \quad\text{where }\;N=\operatorname{lcm}(n_1,\dots,n_k).$$ Consider a prime $p$ not dividing $N$. Can $\dfrac1p$ be a polynomial in $\dfrac 1N$?

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