Vakil's FOAG, Exercise 9.2.K: Transcendental Complex Numbers How does one realize a transcendental complex number as a maximal ideal of $\mathbb{Q}(t) \otimes_{\mathbb{Q}} \mathbb{C}$? This is the essence of Exercise 9.2.K in Vakil's FOAG.
Here is what I've got so far: If $\alpha \in \mathbb{C}$ has transcendence degree $1$, then by definition $\mathbb{Q}(\alpha) \simeq \mathbb{Q}(t)[x]/(p(x))$ for some irreducible polynomial $p(x) \in \mathbb{Q}(t)[x]$. But I don't know where to go with this, or whether something similar may work for $\alpha$ with higher transcendence degree. Is there a way to extend this argument to work for higher transcendence degrees?
 A: To show that $\text{Spec}\,\mathbb{Q}(t) \otimes_\mathbb{Q} \mathbb{C}$ has closed points in natural correspondence with the transcendental complex numbers, we use the formula for how tensor product behaves with respect to localization. We have $\mathbb{Q}(t) = \mathbb{Q}[t]$ localized at $S$, where $S = $ the elements of $\mathbb{Q}(t)$ of positive degree, and so\begin{align*} \mathbb{Q}(t) \otimes_\mathbb{Q} \mathbb{C} & = \mathbb{Q}[t] \text{ localized at }S \otimes_{\mathbb{Q}[t]} \mathbb{Q}[t] \otimes_\mathbb{Q} \mathbb{C} \\ & = \mathbb{Q}[t] \text{ localized at }S \otimes_{\mathbb{Q}[t]} \mathbb{C}[t] \\ & = \mathbb{C}[t] \text{ localized at }S,\end{align*}which is what we want.

Here is what I've got so far: If $\alpha \in \mathbb{C}$ has transcendence degree $1$, then by definition $\mathbb{Q}(\alpha) \simeq \mathbb{Q}(t)[x]/(p(x))$ for some irreducible polynomial $p(x) \in \mathbb{Q}(t)[x]$. But I don't know where to go with this, or whether something similar may work for $\alpha$ with higher transcendence degree.
What is $p(x)$, say for $\alpha = \pi$?
I honestly have no idea. By definition, I think, a field extension $K \hookrightarrow K'$ has transcendence degree $1$ if $K'$ is isomorphic to an algebraic extension of $K(t)$.

If $\alpha$ is transcendental, then $\mathbb{Q}(\alpha)$ is just isomorphic to $\mathbb{Q}(t)$. This does not seem like a super useful direction to go in. The fact that we know that $\mathbb{Q}(t)$ is the same as $\mathbb{Q}(\pi)$ does not seem all that useful because we care about the polynomial $t - \pi$ in $\mathbb{Q}(t)$, not $\pi$.

What is the transcendence degree of an element? Can it be anything other than $0$ or $1$?
By the transcendence degree of $\alpha \in \mathbb{C}$ I mean the transcendence degree of the extension $\mathbb{Q} \hookrightarrow \mathbb{Q}(\alpha)$. I have no idea if can be anything other than $0$ or $1$.

The transcendence degree of an element $\alpha$ can not be anything other than $0$ or $1$. It is $1$ if and only if $\alpha$ is transcendental because if $\alpha$ is algebraic, it is a finite-dimensional vector space over $\mathbb{Q}$, and if it is transcendental, $\mathbb{Q}(\alpha)$ is isomorphic to $\mathbb{Q}(t)$.
