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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!

Community
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amiz9
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    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) – Adam Hughes Jun 02 '16 at 01:23
  • 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 $n \neq 0$? – amiz9 Jun 02 '16 at 01:38
  • 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. – Adam Hughes Jun 02 '16 at 01:41
  • 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 – amiz9 Jun 02 '16 at 01:45
  • Adam Hughes While that is true, it isn't that trivial to show; just because they are pairwise linearly independent doesn't mean that an arbitrary collection will always be linearly independent. – Ege Erdil Jun 02 '16 at 01:58
  • @amiz9 no, I said it was infinite, reread my original comment. – Adam Hughes Jun 02 '16 at 02:02
  • @Starfall he has all of $\sqrt{\Bbb N}$ this includes, for example, all primes, of which there are infinitely many and all of them are quite square free. One can easily Galois the linear independence if need be, or more basic things such as discriminants also do the trick – Adam Hughes Jun 02 '16 at 02:03
  • And I am saying that you have not proven that the square roots of square free integers are linearly independent over $ \mathbb{Q} $, you have only proven that any two of them are. Just because you have that $ v_i, v_j $ are linearly independent for all $ i \neq j $ doesn't mean that the set formed by all of the $ v_i $ is linearly independent. – Ege Erdil Jun 02 '16 at 02:05
  • @Starfall obviously, but I'm not trying to, I'm just commenting with quick and dirty answers because I'm lazy. Also, the site crashed while I was editing my last comment, see that for the elaboration I had been adding. – Adam Hughes Jun 02 '16 at 02:06
  • @AdamHughes ok thanks (sorry I misread), so just to clarify so I understand: $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, ..., \sqrt{n})/\mathbb{Q}$ is an infinite separable field extension of characteristic $0$ of infinite degree? – amiz9 Jun 02 '16 at 02:12
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    @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 $\Bbb Q(\sqrt{1},\ldots, \sqrt{n},\ldots)$ – Adam Hughes Jun 02 '16 at 02:13
  • yes it doesn't 'stop' at $n$ but I was not unsure how to write it, I gues the $...$'s will do, right? – amiz9 Jun 02 '16 at 02:16

2 Answers2

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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.

Andrew Dudzik
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  • 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) – amiz9 Jun 02 '16 at 02:42
  • @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. – Andrew Dudzik Jun 02 '16 at 04:18
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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 $p$.

Noah Schweber
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  • 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 – amiz9 Jun 02 '16 at 01:44
  • @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? – Noah Schweber Jun 02 '16 at 01:50
  • Yes, although I do not know why $\mathbb{F}_p$, (sorry for my ignorance, I find this area really tricky for me) – amiz9 Jun 02 '16 at 01:56
  • @amiz9 Do you understand the definition of $\mathbb{F}_p$? (That is, multiplication and addition modulo $p$.) – Noah Schweber Jun 02 '16 at 02:08
  • oh yes of course I see it now: $\underbrace{1+...+1}_\text{p times}=0$ since $p=0$ mod $p$, right? – amiz9 Jun 02 '16 at 02:14