I have been trying to find examples (and non-examples) of fields which are separable, finite and have characteristic equal to zero.
Separable
Example: $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ because the minimal polynomial $x^2-2$ has $2$ distinct roots.
Non-example: Need to find $L/K$ such that the minimal polynomial has distinct roots. Would $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ work since the minimal polynomial is $(x-\sqrt2)(x-\sqrt{3})$. The poster of this question claims that such a field must be infinite and have characteristic $\neq0$ but I do not see why.
Finite
Example: $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ because $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}]=2 < \infty$.
Non-example: I am not sure. We come up with $L/K$ such that the minimal polynomial of $L$ over $K$ has infinite degree. Would $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, …, \sqrt{n})/\mathbb{Q}$ work, where $n\in\mathbb{N}$? The minimal polynomial in this case would be $(x-\sqrt{2})\cdot … \cdot(x-\sqrt{n})$ which has theoretically infinite degree since $n$ can be made arbitrarily large. (I feel like this is wrong.)
Characteristic equal to zero
Example: Not an extension but the field $\mathbb{Q}$ has characteristic $0$ since the smallest $n$ such that $\sum_{i=1}^n1=0$ is $n=0$ ($1$ is the additive identity in $\mathbb{Q}$.
Non-example: I am not sure what kind of field allows $\sum_{i=1}^n1=0$ for $n \neq 0$. Could you help me please?
So, for short I have $3$ questions:
- What is an example of a non-separable field extension?
- Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, …, \sqrt{n})/\mathbb{Q}$ an infinite field? Edit: yes.
- What fields have characteristic $\neq 0$? Edit: $\mathbb{F_p}(t)/\mathbb{F_p}(t^p)$.
Or even, is there a field extension that satisfies all $3$ properties?
Thank you!
Best Answer
For $p$ a prime, define $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$. Then $\mathbb{F}_p$ is a finite field of characteristic $p$.
The standard example of a non-separable field extension is $\mathbb{F}_p (t^p) \subset \mathbb{F}_p (t)$. The minimal polynomial of $t$ over $\mathbb{F}_p (t^p)$ is $X^p - t^p$, which factors as $(X-t)^p$ over $\mathbb{F}_p (t)$.
To see that every extension $K\subset L$ of a characteristic $0$ field is separable, consider the minimal polynomial $f(X)$ of some algebraic $\alpha\in L$. Because $K$ has characteristic $0$, $f'(X)$ is not the zero polynomial. But if $\alpha$ is a double root of $f$, then $f'(\alpha)=0$, contradicting the assumption that $f(X)$ was minimal.
If $K$ is finite, then $K\subset L$ is always separable as well. To see this, take some algebraic $\alpha\in L$. Then $K\subset K(\alpha)$ is an extension of finite fields. But we can classify all extensions of finite fields, and see that they are all separable.