Infinite field with finite characteristic I need a good example of infinite field with finite characteristic. Can anyone help? 
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Rational functions over a finite field. 
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. 
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 $\displaystyle \mathbb{F}_7.$ 
* 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 $2$. 
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. 
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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. Quote:

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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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 
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