Intereting Posts

measurability of a function -equivalent conditions
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Isomorphic quotient groups $\frac{G}{H} \cong \frac{G}{K}$ imply $H \cong K$?
Why are generating functions useful?
How many combinations can I make?
Angle between 2 points
Show that $e^x > 1 + x + x^2/2! + \cdots + x^k/k!$ for $n \geq 0$, $x > 0$ by induction
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A Counter example for direct summand
Find a solution of the polynomial
Calculating mean and Gaussian curvature
Asymptotic expression for sum of first n prime numbers?

Let $k$ be a field of characteristic $0$ with $\mathrm{trdeg}_\mathbb{Q}(k)$ at most the cardinality of the continuum. I want to prove **the existence of a field homomorphism $k\rightarrow\mathbb{C}$**. (I hope this statement is even true, I made it up on my own.)

Let $S$ be a transcedence base for $k/\mathbb{Q}$, $S’$ one for $\mathbb{C}/\mathbb{Q}$. Let $S\rightarrow S’$ be an injection. The induced map $\mathbb{Q}[S]\rightarrow\mathbb{Q}[S’]\rightarrow\mathbb{Q}(S’)\rightarrow\mathbb{C}$ is injective, and hence (by the mapping property of the fraction field) induces a map $\mathbb{Q}(S)\rightarrow\mathbb{C}$. But as $k/\mathbb{Q}(S)$ is algebraic, we get an induced map $k\rightarrow\mathbb{C}$.

Is this proof correct?

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Your proof seems all right to me.

As an aside, this method can be used to prove that $\mathbb{C}$ admits infinitely many proper subfields isomorphic to itself: simply choose any non-surjective injection $S’\to S’$, which will induce a non-surjective morphism $\mathbb{Q}(S’)\to \mathbb{Q}(S’)$ and thus a non-surjective morphism $\mathbb{C}\to \mathbb{C}$.

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