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The problem asks to find two different algebraically closed fields $\mathcal{E}$ and $\mathcal{F}$ with $\mathcal{E} \subseteq \mathcal{F}$.

We have not done a whole lot of stuff with algebraically closed and in fact the only one that came to mind was $\mathbb{C}$.

Does $\mathbb{C}(x)$, the field of rational functions in $x$ with coefficients from $\mathbb{C}$ work? Our definition of an algebraically closed field (which I'm not sure if there are other standard definitions or not) is that every polynomial with coefficients from the field must have a root in the field. Since the set of polynomials in $\mathbb{C}(x)$ is just $\mathbb{C}[x]$ it seems like my example is algebraically closed as well. Does this in fact work or am I missing something?

Thanks

Edit: I did notice a flaw in my thinking. I need to think of polynomials with coefficients in $\mathbb{C}(x)$ not polynomials contained in $\mathbb{C}(x)$.

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    $\begingroup$ For different, let $E$ be the field of algebraic numbers, and let $F$ be the field of complex numbers. If you want an example related to your suggestion we could use the algebraic closure of $C(x)$, more complicated. $\endgroup$ – André Nicolas Jul 15 '15 at 2:14
  • $\begingroup$ What is the field of algebraic numbers? $\endgroup$ – TuoTuo Jul 15 '15 at 4:44
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The algebraic closure of $\Bbb Q, \overline{\Bbb Q} \subset \Bbb C$ gives an example to your main question.

More generally, for every (edit : u uncountable) cardinality, there is a unique algebraically closed field (upto isomorphism) So while $\Bbb C \subset \overline{C(X)}$(the algebraic closure of $C(X)$) are another pair of examples, the two fields are isomorphic although they are not equal.

(The uniqueness requires the axiom of choice.)

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    $\begingroup$ Thanks for the example. I did not know that the algebraic closure of $\mathbb{Q}$ was strictly smaller than $\mathbb{C}$. $\endgroup$ – TuoTuo Jul 15 '15 at 4:45
  • $\begingroup$ $\overline{\Bbb Q}$ is countable, $\Bbb C$ is not. $\endgroup$ – Asvin Jul 15 '15 at 11:50
  • $\begingroup$ I don't think it is correct to say that for every cardinality there is a unique algebraically closed field (up to isomorphism) of that cardinality. Assuming isomorphism means ring isomorphism and not something like set isomorphism, the statement at least requires that the characteristics of the fields are the same, but even then it doesn't seem correct to me. If you take an algebraic closure of Q(x), are you saying that's abstractly isomorphic to the algebraic closure of Q? $\endgroup$ – CJD Jul 22 '18 at 13:58
  • $\begingroup$ You are absolutely correct, i was mistaken /forgot to mention that you need to assume atleast uncountable cardinality. $\endgroup$ – Asvin Jul 22 '18 at 18:28
  • $\begingroup$ You also need to fix the characteristic of the field. A correct, complete statement is that two algebraically closed fields are isomorphic iff they have the same characteristic and the same transcendence degree over their prime subfield. See e.g. Section 11.2 of math.uga.edu/~pete/FieldTheory.pdf. $\endgroup$ – Pete L. Clark Jul 23 '18 at 15:44
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$\mathbb C(x)$ is not algebraically closed. For example, the polynomial $y^2 - x$ does not have a root in $\mathbb C(x)$.

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  • $\begingroup$ Yeah I was just thinking it was not algebraically closed.. unfortunately. $\endgroup$ – TuoTuo Jul 15 '15 at 2:00
  • $\begingroup$ @billford: I am missing an answer to Andrew's main question...why did you post this as an answer and not as a comment? $\endgroup$ – LSt Jul 15 '15 at 2:11
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    $\begingroup$ @LSt I posted this as an answer because his question is "Does $\mathbb C(x)$ work?" $\endgroup$ – user254385 Jul 15 '15 at 2:23

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