Is it possible to construct a field larger than the complex numbers?

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?

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).

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...