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June 5th, 2017, 06:14 AM   #1
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Infinite field with finite characteristic

I need a good example of infinite field with finite characteristic. Can anyone help?
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June 5th, 2017, 06:19 AM   #2
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I need a good example of infinite field with finite characteristic. Can anyone help?
Algebraic closure of any finite field.
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June 5th, 2017, 07:21 AM   #3
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Rational functions over a finite field.
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June 5th, 2017, 08:01 AM   #4
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I would like you to be more precisive, if that is not a problem. Can you clarify it using a particular field? It will surely become clearer to me.
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June 5th, 2017, 08:39 AM   #5
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The field of all fractions $\displaystyle \frac{f(x)}{g(x)}$ where $\displaystyle f(x),g(x)\in \mathbb{F}_7[x],\ g(x)\neq 0.$

It is the field of rational functions in $\displaystyle x$
$\displaystyle \frac{a_n x^n + a_{n-1} x^{n-1} +\ \ldots\ + a_0}{b_mx^m+b_{m-1}x^{m-1}+\ \ldots\ +b_0}$
with coefficients $\displaystyle a_i, b_j$ in $\displaystyle \mathbb{F}_7.$

Last edited by ABVictor; June 5th, 2017 at 08:43 AM.
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June 5th, 2017, 09:50 AM   #6
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* No finite field is algebraically closed. https://math.stackexchange.com/quest...d-fields-exist

* Every field has an algebraic closure. https://en.wikipedia.org/wiki/Algebraic_closure

Therefore the algebraic closure of, say, $\mathbf F_2$ is an infinite field of characteristic $2$.
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June 5th, 2017, 03:56 PM   #7
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Does have a structure like ? Sorry for all those elementary questions, but I am right after first lecture of field theory, so I can't know much.

Last edited by IAmABread; June 5th, 2017 at 03:59 PM.
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June 5th, 2017, 04:04 PM   #8
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Does $\displaystyle \mathbb{F}_7$ have a structure like $\displaystyle \mathbb{Z}_7$?
Yes, $\mathbf F_p$, the field with $p$ elements ($p$ a prime) is the same as $\mathbb Z_p$, the ring of integers mod $p$.

Where it gets weird though is for the finite fields that are nontrivial powers of $p$. For example $\mathbb Z_4$ is not a field because $2 \times 2 = 0$. But there is a field of order $4$, whose addition and multiplication table you might try working out.

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Sorry for all those elementary questions, but I am right after first lecture of field theory, so I can't know much.
Best to not worry too much about these things at the moment. Your questions are good but the answers aren't elementary. Finite fields are tricky and so is the proof of the existence of algebraic closures.
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June 8th, 2017, 09:28 AM   #9
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Okay, let's say I take coefficients from . What will multiplication identity look like here? Is it just or something more complex? If not, then I assume characteristic is equal 3 according to what I typed. Am I right in all of this?
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June 8th, 2017, 10:34 AM   #10
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Quote:
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Okay, let's say I take coefficients from .
Coefficients of what?


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What will multiplication identity look like here?
Of what ring? You seem to have some hidden assumptions, or you're working in a ring you haven't told us about.

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Originally Posted by IAmABread View Post
Is it just or something more complex?
Is this a ring of formal power series? Where did that come from?

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If not, then I assume characteristic is equal 3 according to what I typed. Am I right in all of this?
No. As far as I can tell your post is missing any context that would make it sensible. Don't mean for that to come out as overly critical. But you lost me at "coefficients." Coefficients of what?

In any event I don't believe the ring of formal power series over a prime-order field is a field. It's an integral domain though.

https://math.stackexchange.com/quest...ntegral-domain

Last edited by Maschke; June 8th, 2017 at 10:52 AM.
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