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This question already has an answer here:

Can you give an example of a field of characteristic zero (other than the complex numbers, real numbers and rational numbers)?

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marked as duplicate by Dietrich Burde, Semiclassical, Alexander Konovalov, user99914, user147263 Apr 26 '16 at 23:33

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    $\begingroup$ You ask lots of questions but accept few answers. Please check to accept when you get an answer that satisfies you. You can also upvote that answer, and others. $\endgroup$ – Ethan Bolker Mar 8 '16 at 15:46
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Rational functions with real coefficients form a field of zero characteristic with usual addition and multiplication:

$$\left\{\frac{a_0 + a_1 x + \dots + a_n x^n}{b_0 + b_1 x + \dots + b_m x^m}\right\}$$

In this field, $\left(\frac{P(x)}{Q(x)}\right)^{-1} = \frac{Q(x)}{P(x)}$

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For any irreducible polynomial $p(x)$ over $\Bbb Q$ (of degree $> 1$), the field $\Bbb Q(\alpha) \cong \Bbb Q[x] / \langle p(x) \rangle$ (where $\alpha$ is a root of $p$) has characteristic $0$ and finite degree $> 1$ over $\Bbb Q$, so it does not coincide with any of $\Bbb Q, \Bbb R, \Bbb C$.

For example, taking $p(x) := x^2 + 1$ gives the field of Gaussian rational numbers, $\Bbb Q[i]$.

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You can consider "A" like the field of the algebraic numbers.

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  • $\begingroup$ What did you mean about algebraic number? $\endgroup$ – user275240 Mar 8 '16 at 15:52
  • $\begingroup$ @user275240 algebraic numbers are roots of rational polynomials. They also form a field of zero characteristic en.wikipedia.org/wiki/Algebraic_number $\endgroup$ – lisyarus Mar 8 '16 at 16:12
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$\mathbb{Q}_p$, the completion of $\mathbb{Q}$ with respect to the $p$-adic absolute value for some prime $p \in \mathbb{Z}$, is a field of characteristic $0$ (it contains $\mathbb{Q}$ as a dense subset). Interestingly (perhaps), the residue field $\mathbb{Z}_p/p\cdot \mathbb{Z}_p \cong \mathbb{F}_p$ has characteristic $p$ and we say $\mathbb{Q}_p$ has mixed characteristic $(0,p)$.

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