Is $Q( (1 + \frac{1}{2})^{2}, (1 + \frac{1}{3})^{3}, (1 + \frac{1}{4})^4, \ldots )$ an algebraic extension of $Q$?

Question

Each of $(1 + \frac{1}{n})^n$ is in $Q$, but is the given extension algebraic? Is the extension just equal to $Q$?

Answer 1

As @Wojowu notes, my specified extension is easily seen to be equal to $Q$ and thus an algebraic extension of $Q$. What follows is a generalization of concepts discussed in the comments above.

Suppose $A$ is a set of elements algebraic over a field $F$. I show that $F(A)$, the smallest field which contains $F$ and $A$ is an algebraic extension of $F$, by showing that $F(A) = \bigcup_{a \in A} F(a)$, and then showing that $\bigcup_{a \in A} F(a)$ is algebraic.

Let $K = \bigcup_{a \in A} F(a)$. If $x \in K$, then $x \in F(a)$ for some $a \in A$, so $x \in F(A)$ and $K \subseteq F(A)$.

Since $F \subseteq K$ and $a \in K$ for each $a \in A$, we have $F(A) \subseteq K$ by minimality. So $K = F(A)$, and K is algebraic because the union of algebraic extensions is an algebraic extension.