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Let $k$ be an algebraically closed field of characteristic 0.

Then I've heard that if $k$ has cardinality no greater than that of $\mathbb{C}$, then there is an embedding $k\hookrightarrow\mathbb{C}$.

Firstly, does anyone have a reference of this statement?

Secondly, suppose $k$ is larger than $\mathbb{C}$ (cardinality-wise). Must there exist an embedding $\mathbb{C}\hookrightarrow k$?

Lastly, suppose again $k$ is larger than $\mathbb{C}$, and let $x\in k$. Must there exist an embedding $\mathbb{C}\hookrightarrow k$ with $x$ in its image?

Does anyone have any references for these facts?

oxeimon
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  • Related: http://math.stackexchange.com/questions/365323/does-every-algebraically-closed-field-contain-the-field-of-complex-numbers –  May 11 '15 at 02:07
  • Please refer to Theorem 4.4 of Pierre A. Grillet’s Abstract Algebra, No. 242, Graduate Texts in Mathematics, Springer, Second Edition, 2007. – Berrick Caleb Fillmore May 11 '15 at 02:08
  • (1) There exists a unique algebraically closed field of every cardinality/characteristic in fact. Don't have a reference offhand unfortunately. (2) Adjoin $2^{\frak c}$-many variables to $\Bbb Q$. The resulting field has no irrational algebraic elements, which precludes any embedding of $\Bbb C$. – anon May 11 '15 at 02:09
  • All three of them are true. They follow from the application of some early results of the theory of trascendental field extensions. Which, along with several others, can be found in Lang, Algebra. –  May 11 '15 at 02:14
  • @whacka I think you must mean to qualify over infinite cardinalites and omit the finite ones. Regards – rschwieb May 11 '15 at 03:02
  • @rschwieb You're sort of right, I meant to specify uncountable cardinalities. :-) – anon May 11 '15 at 03:36
  • @whacka $ Q\bar$ and algebraic closure of $ Q\bar (t)$ have same cardinality (both countable infinite), but they are definitely different... I think you meant cardinality of transcendental basis? – lee Nov 29 '15 at 00:57
  • @lee Did you see my last comment right above yours? – anon Nov 29 '15 at 21:24

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