2
$\begingroup$

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:

  1. What is an example of a non-separable field extension?
  2. Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, ..., \sqrt{n})/\mathbb{Q}$ an infinite field? Edit: yes.
  3. 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!

$\endgroup$
12
  • 1
    $\begingroup$ All characteristic $0$ extensions are separable. To see this, note that the derivatives of a non-constant polynomial cannot vanish identically in characteristic $0$. Adjoining infinitely many square roots does produce an infinite degree extension. There are tons of fields of non $0$ characteristic (infinitely many) $\endgroup$ Jun 2, 2016 at 1:23
  • $\begingroup$ Why does it not produce an infinite degree extension if the degree of the minimal polynomial is arbitrarily large? What would our additive identity, $1_{id}$ need to be such that $n(1_{id})=0$ for $n \neq 0$? $\endgroup$
    – amiz9
    Jun 2, 2016 at 1:38
  • $\begingroup$ You just show that it has infinite dimensions as a vector space, you know $\sqrt{d}, \sqrt{d'}$ are linearly independent for square free integers, $d$, so $1,\sqrt{d_1},\ldots$ is an infinite set which is linearly independent, hence the dimension as a vector space is infinite. $\endgroup$ Jun 2, 2016 at 1:41
  • $\begingroup$ I thought you said adjoining infinitely many square roots does NOT produce an infinite degree extension.. but you say they are are linearly independent infinite set? Or are you constructing a different example to mine? Thanks $\endgroup$
    – amiz9
    Jun 2, 2016 at 1:45
  • 2
    $\begingroup$ @amiz9 oh, if you're stopping at $\sqrt{n}$ it isn't, I thought you had just forgotten to keep going after that. If you stop at $\sqrt{n}$ your basis has size bounded by $2^n$ by the tower laws. I was referring to $\Bbb Q(\sqrt{1},\ldots, \sqrt{n},\ldots)$ $\endgroup$ Jun 2, 2016 at 2:13

2 Answers 2

10
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ In fact is it true that $\mathbb{F_p}(t)/\mathbb{F_p}(t^p)$ is infinte, non-separable and has non-zero characteristic? (I believe so) $\endgroup$
    – amiz9
    Jun 2, 2016 at 2:42
  • $\begingroup$ @amiz9 There are two fields. They are both infinite, with non-zero characteristic, and the extension from one to the other is not separable. The field extension has finite degree. $\endgroup$ Jun 2, 2016 at 4:18
1
$\begingroup$

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 $p$.

$\endgroup$
5
  • $\begingroup$ So with the field of rational functions over $\mathbb{F}_p$, why does $\underbrace{1+...+1}_\text{p times}=0$? What is the identity in this case? Thanks $\endgroup$
    – amiz9
    Jun 2, 2016 at 1:44
  • $\begingroup$ @amiz9 Well, remember that "$1$" and "$0$" in this context mean $1$ and $0$ in $\mathbb{F}_p$ (think about how any field sits inside its field of rational functions), so $\mathbb{F}_p(x)$ has characteristic $p$ exactly because $\mathbb{F}_p$ does. Does that make sense? $\endgroup$ Jun 2, 2016 at 1:50
  • $\begingroup$ Yes, although I do not know why $\mathbb{F}_p$, (sorry for my ignorance, I find this area really tricky for me) $\endgroup$
    – amiz9
    Jun 2, 2016 at 1:56
  • $\begingroup$ @amiz9 Do you understand the definition of $\mathbb{F}_p$? (That is, multiplication and addition modulo $p$.) $\endgroup$ Jun 2, 2016 at 2:08
  • $\begingroup$ oh yes of course I see it now: $\underbrace{1+...+1}_\text{p times}=0$ since $p=0$ mod $p$, right? $\endgroup$
    – amiz9
    Jun 2, 2016 at 2:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .