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The third statement to prove would be "Prove that $\mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Q}$-algebras"

I thought about proving those statements in the following way:

$\mathbb{C}$ is a finite $\mathbb{R}$-algebra

The homomorphism we use is h : $\mathbb{R}$ -> $\mathbb{R}$ and h = $id_{\mathbb{R}}$, and once we have h($\mathbb{R}$) (which is $\mathbb{R}$ itself...) we can say that $\mathbb{C}$ = h($\mathbb{R}$)[i] (because we only have to add the element i to our set h($\mathbb{R}$) and the smallest ring containing h($\mathbb{R}$) and i is $\mathbb{C}$)

$\mathbb{Q}, \mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Z}$-algebras

Because $\mathbb{Q}$ ⊂ $\mathbb{R}$ ⊂ $\mathbb{C}$ we only do it for $\mathbb{Q}$

There does not exist a one argument homomorphism which gives you any rational number you want, we can only work with some kind of homomorphism which gives you an integer. So to make $\mathbb{Q}$ out of it, we'd have to add an infinite amount of elements/numbers (all the "fully" rational ones like $\frac{1}{2}$, ...) to our set h($\mathbb{Z}$) to get $\mathbb{Q}$, but that's not what we're looking for. thus it's not a finite $\mathbb{Z}$-Algebra

$\mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Q}$-algebras

Because $\mathbb{R}$ ⊂ $\mathbb{C}$ we only do it for $\mathbb{R}$

Same understanding as above, there does not exist a one argument homomorphism which gives you any irrational number you'd like (and there's an infinite amount of them), and working with a homomorphism which gives you a rational number as a result, the only way to create $\mathbb{R}$ out of h($\mathbb{Q}$) is to add all the irrational numbers to the set h($\mathbb{Q}$), but that's not what we're looking for

Is this enough? Because I kind of feel that it's too little, and that some sort of mathematical explanation is lacking (some equations for example)

user
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  • One of the biggest problems with what you're doing is that it seems like you're assuming that subalgebras of finitely generated algebras are also finitely generated. That is not true.. I'm not aware of a special exception for $\mathbb Z$, and I'm not saying there isn't a valid argument along these lines, but one should say something more to justify the lines that say "we can just do it for " – rschwieb Mar 08 '21 at 14:16
  • You mean $\mathbb{Q}$ ⊂ $\mathbb{R}$ ⊂ $\mathbb{C}$ we only do it for $\mathbb{Z}$? Sry it should be $\mathbb{Q}$ ⊂ $\mathbb{R}$ ⊂ $\mathbb{C}$ we only do it for $\mathbb{Q}$. But in the end, we could also do the same reasoning for $\mathbb{R}$ and $\mathbb{C}$, that you can't create a homomorphism starting from $\mathbb{Z}$ and expand its image by a finite amount of elements to get $\mathbb{R}$ and $\mathbb{C}$. I mean it's kind of trivial and what I'm asking is if I need something more mathematically to show it – user Mar 08 '21 at 14:55
  • See also https://math.stackexchange.com/questions/148745/fields-finitely-generated-as-mathbb-z-algebras-are-finite – lhf Mar 08 '21 at 15:05

1 Answers1

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$\mathbb{Q}, \mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Z}$-algebras

Hint for $\mathbb R$ and $\mathbb C$:
A finitely generated $\mathbb Z$-algebra $A$ is the image of a ring homomorphism $\mathbb Z[x_1,\dots,x_n] \to A$ and therefore is countable.

Hint for $\mathbb Q$:
The numbers in $\mathbb Z[q_1,\dots,q_n]$ have denominators whose prime divisors come from a finite set.

lhf
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  • So because it is countable but R and C are uncountable, we cannot extent the image with a finite set of elements to get R and C?
    • – user Mar 09 '21 at 15:45
  • What are prime divisors? I don't really see how we should get the solution here
    • – user Mar 09 '21 at 15:47
  • @user: 1: because the image is countable it cannot be $\mathbb R$ or $\mathbb C$; 2. The prime divisors of the denominators of the $q_i$ form a finite set. – lhf Mar 09 '21 at 15:49
  • "The prime divisors of the denominators of the form a finite set." so how can I use this to show that Q is not a Z-algebra? I don't see the link here – user Mar 09 '21 at 15:53