Let $t$ be a transcendental number. Prove that the set $\{(a+bt) \mid a,b \in \mathbb{Q}\}$ is not a number field.

Question

Can I just pick a number in the set and then prove it's not constructible? Thx

Answer 1

By definition a number field $K$ is a finite extension of $\mathbb{Q}$, so in particular, every element of a number field is algebraic, thus no transcendental number can be an element of a number field.

This is why number fields are also called algebraic number fields, since their elements are algebraic numbers ;-)

Answer 2

Here's something worth noting: if $\{(a+bt) \mid a,b \in \mathbb{Q}\}$ with $t$ transcendental over $\Bbb Q$ were a field of any sort, then $(a + bt)^2 = a^2 + 2abt + t^2 = c + dt$ for some $c, d \in \Bbb Q$. Then $t^2 + (2ab - d)t + (a^2 - c) = 0$; but this implies the the degree of $t$ over $\Bbb Q$ is at most $2$; $t$ transcendental? NOT!!!

It appears this argument can be extended to cover the case $\{\sum_0^n a_i t^i \mid a_i \in \Bbb Q \}$ with $t$ transcendental as well, and maybe even applied to some other base fields. Something to think about in your copious spare time!

Hope this helps. Cheers,

and as always,

Fiat Lux!!!