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If $F=\mathbb{Q}(u_{1},…,u_{n})$ is a finitely generated extension of $\mathbb{Q}$,to proof exists an embedding $\varphi:F\longrightarrow \mathbb{C}/\mathbb{Q}$.

I'm actually don't know this question's background.I'm not exactly sure how to start this problem. Any help at all would be a tremendous help.

I am a student from China.I am learning Abstract Algebra in this term.It's my first time to ask question on this website,hope you can give me some valuable advice.

Matrix
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  • Welcome to the site! I need to begin by referring you to our abridged guide for new askers. A problem with your question is that your question looks like an attempt to outsource a homework assignment, which is something we frown upon, unless you can add more context to it (see the linked thread for tips). Observe that for the purposes of the rule it is irrelevant whether this actually is a homework question or something that you found in a book. The rule applies. – Jyrki Lahtonen Dec 03 '24 at 07:23
  • Another thing is that the question is quite natural. Meaning that we may have covered it on the site already. Searching and finding a good match need not be easy, but you should be prepared for the question to be a duplicate of an earlier one. – Jyrki Lahtonen Dec 03 '24 at 07:24
  • For example, I could answer this easily, but to do so in a helpful way I would need to know quite a bit about your background. How familiar you are with concepts such as: transcendence degree, algebraically closed field and such? Such details can become useful context in the question body. – Jyrki Lahtonen Dec 03 '24 at 07:30
  • Gotta commute. May be you get more help in the meantime? Good luck with this site, it's a wonderful resource. – Jyrki Lahtonen Dec 03 '24 at 07:31
  • @JyrkiLahtonen I am a Chinese student.I am learning Abstract Algebra in this term,the concept you mentioned I alrea have learned. – Matrix Dec 03 '24 at 07:39
  • All polynomial in $\mathbb{Q}$ has a zero in $\mathbb{C}$. That is, algebraic closure of $\mathbb{Q}$ is embedded inside $\mathbb{C}$. Try to construct $F\to \mathbb{C}$ using this fact. – J1U Dec 03 '24 at 07:45
  • @J1U sorry,can you give me the specific mapping relationship? – Matrix Dec 03 '24 at 10:31
  • Let me just give an easy example. $\mathbb{Q}(i)=\mathbb{Q}[x]/(x^2+1)$ is an algebraic extension over $\mathbb{Q}$. $x^2+1$ has zeroes $i$ and $-i$ in $\mathbb{C}$. Now it is obvious how $\mathbb{Q}[x]/(x^2+1)\to\mathbb{C}$ is defined: map $x$ to either $i$ or $-i$. – J1U Dec 03 '24 at 11:05
  • Ok, it is a good idea to add the pieces of information about your background to the question body. Reviewers will not necessarily see it here in the comment section. – Jyrki Lahtonen Dec 03 '24 at 13:01

2 Answers2

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With basic facts about transcendence degrees in place we can argue as follows.

The extension $F/\Bbb{Q}$ is finitely generated, so its transcendence degree $\ell$ is finite and $\ell\le n$. Let us fix a transcendence basis $\{v_1,v_2,\ldots,v_\ell\}$. For its part $\Bbb{C}/\Bbb{Q}$ is known to have infinite transcendence degree. This follows from a cardinality argument, but using the Lindemann-Weierstrass theorem we can more explicitly state that the numbers $z_1=e^{\sqrt2}$, $z_2=e^{\sqrt3}$, $\ldots, z_\ell=e^{\sqrt{p_\ell}}$ are all transcendental and algebraically independent. Here the exponents are the square roots of the $\ell$ smallest primes.

This means that the subfield $K=\Bbb{Q}(v_1,v_2,\ldots,v_\ell)\subseteq F$ is isomorphic to the subfield $\Bbb{Q}(z_1,z_2,\ldots,z_\ell)\subseteq \Bbb{C}$ by an embedding of fields $\phi':K\to\Bbb{C}$ fully determined by $v_i\mapsto z_i, i=1,2,\ldots,\ell$.

By the properties of transcendence bases we know that $F/K$ is algebraic and finitely generated. Therefore actually $[F:K]$ is finite. As $\Bbb{C}$ is algebraically closed, a textbook lemma I described here shows that we can then extend $\phi'$ to an embedding $\phi:F\to\Bbb{C}$ — adding one generator to the domain at a time. First extending it to an embedding of $F_1=K(u_1)\to\Bbb{C}$, then to an embedding of $F_2=F_1(u_2)$ etc. Observe that all the extensions $F_{i+1}/F_i$ are simple and algebraic.

Jyrki Lahtonen
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An alternative way to look at it:

It is enough to find an injective map $$\mathbb{Q}[u_1, \ldots, u_n]\to \mathbb{C}$$ Consider the canonical map from the ring of polynomial in $n$ variables $x_i$, $i=1,n$ to our domain $$\mathbb{Q}[x_1, \ldots, x_n]\to \mathbb{Q}[u_1, \ldots, u_n]$$ This map has a kernel $\mathcal{P}$, a prime ideal of $\mathbb{Q}[x_1, \ldots, x_n]$. Now, a map $\mathbb{Q}[u_1, \ldots, u_n]\to \mathbb{C}$ will be given by an assignment $u_i \to t_i$, $t_i \in \mathbb{C}$ such that $t=(t_1, \ldots, t_n)$ satisfies all of the equations $P(t)=0$, where $P\in \mathbb{P}$. Note that since $\mathbb{P}$ is finitely generated ( Hilbert), we may consider only finitely many equation $P_1$, $\ldots$, $P_m$. In other words, a map $\mathbb{Q}[u_1, \ldots, u_n]\to \mathbb{C}$ is given by a choice of a point $t$ in the zero set $Z$ in $\mathbb{C}^n$ of the ideal $\mathcal{P}$. Now, when is the map given by $t$ injective? If and only if $t$ satisfies no other polynomial equations over $\mathbb{Q}$ than the ones in $\mathbb{P}$. In other words, an injective map $\mathbb{Q}[u_1, \ldots, u_n]\to \mathbb{C}$ is given by choosing a "generic point" in the zero set $Z$. Now, if we take a "random point" in $Z$, how often will that be generic? Almost always, and this could be argued with measure theory, since non-generic points lie on a countably union of lower-dimensional subvarieties of $Z$, and thus form a subset of $0$ measure.

orangeskid
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