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Can we extend the complex numbers in any way such that $\mathbb{C} \subset\mathbb{C}[a]$ ? Or is $\mathbb{C}$ the extension to end all extensions?

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    $\begingroup$ Surcomplex numbers $\endgroup$ – user137731 Oct 5 '15 at 22:51
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    $\begingroup$ Sure, construct the field $\mathbb{C}(z)$ of rational functions in $z$ with complex coefficients. $\endgroup$ – James Oct 5 '15 at 22:52
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    $\begingroup$ en.wikipedia.org/wiki/Quaternion ? (skew field) $\endgroup$ – Clement C. Oct 5 '15 at 22:55
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    $\begingroup$ @ClementC. The Quaternions arn't commutative, so they aren't a field. $\endgroup$ – Kyle Oct 5 '15 at 22:56
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    $\begingroup$ Yes, I'm still deeply puzzled about that in English (in French, commutativity is optional for fields). I mentioned it's a skew field in my answer, as I was unsure about how much the OP cared about commutativity. $\endgroup$ – Clement C. Oct 5 '15 at 22:57
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Given any field, we can always construct bigger fields. If our field isn't algebraically closed, we can adjoin new roots of polynomials, otherwise we can adjoin transcendental elements (which is equivalent to forming a field of rational functions). In fact, every field extension is an algebraic extension of a purely transcendental extension.

(As $\Bbb C$ is algebraically closed, it has no algebraic extensions, so no finite ones.)

In particular, $\Bbb C(T)$ (the field of rational functions in the variable $T$ with complex coefficients) is bigger than $\Bbb C$ in the sense of set-theoretic inclusion. However, the algebraic closure has the same cardinality as $\Bbb C$ itself, and therefore is abstractly isomorphic to $\Bbb C$, which means there exists a way to embed $\Bbb C(T)$ inside $\Bbb C$. If we want to go bigger in the sense of cardinality, we can form the field of complex-coefficient rational functions in $\kappa$-many variables, where $\kappa$ is a cardinal number bigger than the continuum $\mathfrak{c}=|\mathbb{C}|$. This is certainly bigger!

Note that there is no subring of the polynomial ring $\mathbb{C}[T]$ that is a field and which strictly contained $\Bbb C$ itself, for if there were then it would contain some nonconstant $f(T)$, hence contain the nonpolynomial element $f(T)^{-1}$, which is impossible inside $\Bbb C[T]$.

Note also that fields of different characteristic are incompatible: fields of different characteristic can never be contained inside a common field. So not only is there not a field to end all fields, but the "class of all fields" juts out in different, mutually exclusive directions (one for each prime, and zero).

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    $\begingroup$ Wait how do you embed $\mathbb{C}(T)$ inside $\mathbb{C}$? $\endgroup$ – Brandon Thomas Van Over Oct 5 '15 at 23:04
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    $\begingroup$ @SirJective The axiom of choice implies $\overline{\Bbb C(T)}$ (the algebraic closure of $\Bbb C(T)$) is isomorphic to $\Bbb C$ itself, say via an isomorphism $\phi:\overline{\Bbb C(T)}\to\Bbb C$. (In a given characteristic, there is a unique algebraically closed field of every infinite cardinality.) Since we require choice, there is no way to construct it explicitly. Then the restriction of $\phi$ to $\Bbb C(T)$ gives an embedding $\Bbb C(T)\to\Bbb C$. $\endgroup$ – whacka Oct 5 '15 at 23:07
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    $\begingroup$ @SirJective you can somehow visualize the embedding by thinking about the complex number not as a topological space, but just a field. Then, an embedding of $\mathbb{C}(T) \to \mathbb{C}$ leads to choosing a well ordered sequence of trascendental extensions of $\mathbb{Q}$ (so a basis of algebricalli independent $(x_0, \dots)$) and then map $T \to x_0$ and $x_n \to x_{n+1}$ for finite $n$. AC is only needed to provide a basis for the extensions, not to create the embedding. $\endgroup$ – rewritten Oct 5 '15 at 23:45
  • $\begingroup$ @rewritten Out of curiosity, how do you create the embedding without a transcendence basis? $\endgroup$ – whacka Oct 6 '15 at 14:12
  • $\begingroup$ @whacka you got me ;) What I mean is that you need AC to find a basis, but once you have a basis (and a countable subset of it - weak AC again), the embedding is easy, and you don't need anything as strong to do it. $\endgroup$ – rewritten Oct 6 '15 at 14:28
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Yet another form: ultraproducts. The cartesian product of fields $$P = {\Bbb C}\times{\Bbb C}\times\cdots$$ isn't a field because has zero divisors: $$(0,1,0,1,\cdots)(1,0,1,0\cdots)=(0,0,0,0,\cdots).$$ But a quotient will be a field. Let be $\mathcal U$a nonprincipal ultrafilter on $\Bbb N$. Define $$(a_1,a_2,\cdots)\sim(b_1,b_2,\cdots)$$ when $$\{n\in\Bbb N\,\vert\, a_n=b_n\}\in\mathcal U.$$ The quotient $F = P/\sim$ is a field strictly bigger (in the sense of cardinality) than $\Bbb C$ and the inclusion $\Bbb C\longrightarrow F$ is...

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If $K$ is any field, the ring of rational functions over $x$ is also a field as is the ring of formal Laurent series over $K$. So there is no field to end all fields.

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