Non-separable, infinite field extensions of non-zero characteristic 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!
 A: 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.
A: You ask, among other things, for an example of a field with characteristic $\not=0$. 
The standard examples are - for $p$ prime - the fields $\mathbb{Z}/p\mathbb{Z}$ of integers modulo $p$ (call these "$\mathbb{F}_p$" for simplicity). 
A couple remarks:


*

*Why did I need $p$ to be prime? Well, let's look at $\mathbb{Z}/6\mathbb{Z}$ for example. This is a ring - I can add and multiply elements - but it's not a field, because $[2]\cdot [3]=[0]$, and so neither $[2]$ nor $[3]$ has an inverse. More generally, $\mathbb{Z}/k\mathbb{Z}$ is a field iff $k$ is prime.

*That said, for $p$ prime, $\mathbb{F}_p$ isn't the only field of characteristic $p$! There are infinite fields of finite characteristic - e.g., the field of rational functions over $\mathbb{F}_p$ - and even finite examples: although $\mathbb{Z}/p^n\mathbb{Z}$ isn't a field for $p$ prime and $n>1$, there is a field of size exactly $p^n$! Building it is somewhat tricky, though.

*Finally, later on in your studies you might see some things called the $p$-adics - these do not have characteristic $p$. 
