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June 5th, 2017, 06:14 AM  #1 
Newbie Joined: Apr 2017 From: Europe Posts: 16 Thanks: 1  Infinite field with finite characteristic
I need a good example of infinite field with finite characteristic. Can anyone help?

June 5th, 2017, 06:19 AM  #2 
Senior Member Joined: Aug 2012 Posts: 2,311 Thanks: 706  
June 5th, 2017, 07:21 AM  #3 
Member Joined: May 2017 From: Russia Posts: 34 Thanks: 5 
Rational functions over a finite field.

June 5th, 2017, 08:01 AM  #4 
Newbie Joined: Apr 2017 From: Europe Posts: 16 Thanks: 1 
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.

June 5th, 2017, 08:39 AM  #5 
Member Joined: May 2017 From: Russia Posts: 34 Thanks: 5 
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 $\displaystyle x$ $\displaystyle \frac{a_n x^n + a_{n1} x^{n1} +\ \ldots\ + a_0}{b_mx^m+b_{m1}x^{m1}+\ \ldots\ +b_0}$ with coefficients $\displaystyle a_i, b_j$ in $\displaystyle \mathbb{F}_7.$ Last edited by ABVictor; June 5th, 2017 at 08:43 AM. 
June 5th, 2017, 09:50 AM  #6 
Senior Member Joined: Aug 2012 Posts: 2,311 Thanks: 706 
* No finite field is algebraically closed. https://math.stackexchange.com/quest...dfieldsexist * 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 $2$. 
June 5th, 2017, 03:56 PM  #7 
Newbie Joined: Apr 2017 From: Europe Posts: 16 Thanks: 1 
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. 
June 5th, 2017, 04:04 PM  #8  
Senior Member Joined: Aug 2012 Posts: 2,311 Thanks: 706  Quote:
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. 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.  
June 8th, 2017, 09:28 AM  #9 
Newbie Joined: Apr 2017 From: Europe Posts: 16 Thanks: 1 
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?

June 8th, 2017, 10:34 AM  #10  
Senior Member Joined: Aug 2012 Posts: 2,311 Thanks: 706  Coefficients of what? Of what ring? You seem to have some hidden assumptions, or you're working in a ring you haven't told us about. Is this a ring of formal power series? Where did that come from? Quote:
In any event I don't believe the ring of formal power series over a primeorder field is a field. It's an integral domain though. https://math.stackexchange.com/quest...ntegraldomain Last edited by Maschke; June 8th, 2017 at 10:52 AM.  

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