I have $2$ similar questions :
- Is the following set $K$ a field: $$K=\underset{i=even}{\bigcup} \mathbb{F}_{2^{i}}$$ where $\mathbb{F}_{2^{i}}$ values are all finite fields?
Is so, what is the characteristic of this field?
- Is $K \cup \mathbb{F}_{2^{3}}$ a field?
My answer is: yes both $1$ and $2$ are infinite fields by subfield criterion of finite fields, but I am not sure!
And what about the characteristic? Is it $2,$ which is the base prime, or is it $0,$ since it is infinite?
Your $K$ is a field, but this is not as trivial as it looks. The non trivial point is to show stablility by $+$.
Let $x\in\mathbb{F}_{2^i}$ and $y\in\mathbb{F}_{2^j}$.
Then $i\mid ij$ and $j\mid ij$, so $\mathbb{F}_{2^i}$ and $\mathbb{F}_{2^j} $ are both contained in $\mathbb{F}_{2^{ij}}$, and $x+y\in \mathbb{F}_{2^{ij}}\subset K$ (since $ij$ is also even).
The characteristic of $K$ is also equal to the characteristic of any subfield, since the unit elements are the same. Since $\mathbb{F}_2$ is a subfield, the characteristic of $K$ is $2$.
Since $\mathbb{F}_{2^3}\subset \mathbb{F}_{2^6}$, your field in Question 2. is just $K$....