Does there exist a field extension of the algebraic numbers $\mathbb{A}$ other than $\mathbb{C}$?

Question

Does there exist a field extension of the algebraic numbers $\mathbb{A}$ other than $\mathbb{C}$?

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I tried constructing one by adding $e$ to the algebraic numbers. We immediately get the group $\{a+eb; (a,b) \in \mathbb{A}^2\}$ but its easy to prove that this is not a field. So the extension must contain numbers like $\frac{1}{c+de}$, but these numbers are not closed under addition; when you add them, you get fractions where the numerators and denominators are polynomials in $e$ with coefficients in $\mathbb{A}$.

So a natural proposition is to consider the set $\mathbb{A}(e)$, i.e the set of rational fractions of polynomials in $\mathbb{A}[x]$ evaluated at $e$. Its now easy to see that this is a field.

We can also verify that $\mathbb{A}(e)$ is a proper subfield of $\mathbb{C}$. Since $\mathbb{A}$ is countable, $\mathbb{A[x]}$ is too, and hence $\mathbb{A}(x)$ is countable as well since its the same thing as $\mathbb{A[x]}^2$. Thus $\mathbb{A}(e)$ is countable, and thus obviously not equal to $\mathbb{C}$.

Unfortunately, the degree of this extension, i.e $[\mathbb{A}(e):\mathbb{A}]$ is infinite. To prove this, suppose this space admits a basis with $n$ elements: $\{f_1(e),..f_n(e)\}$ where $f_1(x),...,f_n(x)$ are elements of $\mathbb{A}(x)$. Let $\{\alpha_1,...,\alpha_m\}$ be the set of poles of $f_1(x),...,f_n(x)$. Hence any linear combination of $f_1(x),...,f_n(x)$ has all of its poles lying in this set. Hence every element in $\mathbb{A}(e)$ can be written as $P(e)/[(e-\alpha_1)...(e-\alpha_m)]$ for some polynomial $P \in \mathbb{A}[x]$. Now take $a \in \mathbb{A}$ not lying in the set $\{\alpha_1,...,\alpha_m\}$. By the above, there exists $P \in \mathbb{A}[x]$ such that

$1/(e-a)=P(e)/[(e-\alpha_1)...(e-\alpha_m)]$.

So $e$ is a solution to $P(x)(x-a)-(x-\alpha_1)...(x-\alpha_m)]=0$. Since the LHS is a polynomial in $\mathbb{A}[x]$, all of its roots are algebraic and thus $e$ can only be a root if its the zero polynomial. But the LHS evaluated at $x=a$ is nonzero, a contradiction.

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I'm not sure if what I wrote above is correct. Assuming it is, my question is what are some other examples of field extensions of the algebraic numbers $\mathbb{A}$ other than $\mathbb{C}$ and $\mathbb{A}(e)$? In particular, are there any finite extensions?

Note: As Thomas Andrews noted, the argument above generalizes to $\mathbb{A}(\alpha)$ for any transendental $\alpha$. What I'm looking for is other examples...

Note 2: I have no background in field theory I just learned what is a field extension this morning, so forgive me if what I wrote above contains some badly wrong statements.

Answer 1

There are no finite extensions, because any such extension is algebraic.

There are however lots of infinite extensions.

For instance for any set $X$ of indeterminates you have the field $\mathbb A(X)$, which is the field of rational functions with variables picked from $X$ and coefficients in $\mathbb A$.

Note that $X$ can have any cardinality, giving you fields bigger than $\mathbb C$.

Answer 2

Formalizing, perhaps, a little more your work: if $\;K\;$ is a finite extension of $\;\Bbb A\;$, then for any $\;k\in K\;$ we get that $\;[\Bbb A(k):\Bbb A]<\infty\;$ , and thus there exists non zero

$$p(x)=a_nx^n+\ldots+a_1x+a_0\in\Bbb A[x]\;,\;\;\text{s.t.}\;\;p(k)=0$$

But let us now take a look at $\;\Bbb Q(k, a_0,...,a_n)\;$ . This is a finite extension of $\;\Bbb Q(a_0,...,a_n)\;$ , and this last is a finite extension of $\;\Bbb Q\;$ as $\;a_i\;$ algebraic for all $\;i\;$ . We thus get $\;\Bbb Q(k, a_0,...,a_n)\;$ is algebraic over $\;\Bbb Q\;$ and thus so is $\;k\;$, from which we get $\;k\in\Bbb A\implies K=\Bbb A\;$ .