# $\mathbb Q$ is not finitely generated $\mathbb Z$-algebra [duplicate]

I proved that $\mathbb Q$ is not finitely generated as $\mathbb Z$ module using Nakayama Lemma. Now I want to prove the following:

$\mathbb Q$ is not finitely generated as $\mathbb Z$- algebra.

I don't want to use the result that: A field which is finitely generated $\mathbb Z$-algebra is finite.

Any help will be appreciated. Many thanks.

## marked as duplicate by user26857 commutative-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 22 '16 at 19:51

• – user26857 Sep 22 '16 at 19:56

Suppose it were. What primes can appear as denominators in any algebra element of $\mathbb Q$? Can there be infinitely many?

In fact, suppose we have a domain $D$ for which $F=Q(D)$ is a finitely generated $D$-algebra. By taking a common denominator, it follows that $F = D[s^{-1}]$ for some $s\in D$. It is evident from this that $\mathbb Q$ is not a finitely generated $\mathbb Z$-algebra, since $s$ can contain only finitely many primes, as suggested above.

Domains $D$ for which $F=Q(D)$ is a finitely generated $D$-algebra are called Goldstein domains. They play an important role in a generalization of Hilbert's Nullstellensatz to what are known as Jacobson (or Hilbert) rings.

If $R$ is any (commutative) ring, a prime $\mathfrak p$ in $R$ is Goldstein if the domain $R/\mathfrak p$ is Goldstein. A Hilbert ring $R$ is a ring all whose Goldstein prime ideals are maximal -- this equivalent to $R$ having all its quotients with nilradical equal to its Jacobson radical.

If $R$ is a Hilbert ring and $S$ is a finitely generated $R$-algebra, then

• $S$ is also a Hilbert ring,
• If $\mathfrak n$ is a maximal ideal in $S$, its contraction $\mathfrak m$ to $R$ is maximal in $R$ and,
• The field extension $\dfrac{S}{\mathfrak n} / \dfrac{R}{\mathfrak m}$ is finite.

In particular taking $k=R$ a field, $k$ is trivially a Hilbert ring, and the above says that if $K$ is a finitely generated algebra that is a field then $[K:k]$ is finite -- this is the content of the Zariski lemma, crucial to proving Hilbert's Nullstellensatz.

Add. I should have noted that the integers are a Jacobson ring, and hence the fact you don't want to use is a consequence of this generalized Nullstellensatz.

• So then $\mathbb Q= \mathbb Z [1/p_1,\ldots,1/p_n],$ where $p_i's$ are primes in $\mathbb Z.$ But then what contradicts it ? – user371231 Sep 22 '16 at 18:11
• Oh i understand , then $1/q$ cannot be an element in $\mathbb Z[1/p_1,\ldots,1/p_n]$ when $q$ is a prime in $\mathbb Z$ different from the $p_i's.$ Am i right ? – user371231 Sep 22 '16 at 18:14
• Yes, you are. ${}{}$ – Pedro Tamaroff Sep 22 '16 at 18:43