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My textbook says that $ (\mathbb{Z}_{m},+,*)$ is a field if and only if m is a prime number.

However, on Wikipedia it says: "Finite fields only exist when the order (size) is a prime power $p^{k}$ (where $p$ is a prime number and $k$ is a positive integer)."

Since 2 is a prime number, this would imply that $ (\mathbb{Z}_{4},+,*)$ is a finite field, since $2^{2} =4$

The same book then demonstrates that $ (\mathbb{Z}_{4},+,*)$ is not a field.

So is $ (\mathbb{Z}_{4},+,*)$ a field or not?

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    $\begingroup$ There is a field with four elements, and the ring $\Bbb Z/4\Bbb Z$ is not a field. These facts are not contradictory. $\endgroup$ – anon May 18 '15 at 16:44
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$\mathbb{Z}_4$ is not a field. There exist fields with prime power orders with exponents other than $1$. However, they are not of the form $\mathbb{Z}_m$ for any $m$. Rather, they are obtained from $\mathbb{Z}_p$ by adjoining roots of polynomials.

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As @MattSamuel indicated, it is not a field.

But be careful.

Just because there is a field of order $p^k$, it doesn't follow that such a field needs to be $\mathbb Z_{p^k}$. It just has the same size as $\mathbb Z_{p^k}$.

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The fields $\Bbb Z_p$, for $p$ a prime, are special, their additive groups are cyclic. For a finite field of order $p^k$, where $k > 1$, their additive groups are not cyclic, because no element has additive order greater than $p$ (this $p$, by the way, is called the characteristic of the field, and can be shown to necessarily be prime, for a finite field). So, for example, in the finite field of order $4\ (= 2^2)$, we have:

$\alpha + \alpha = 0$, for all $\alpha \in \Bbb F_4$ (it's a field of characteristic $2$),

that is, every non-zero element has additive order $2$.

Contrast this with $\Bbb Z_4$, where both $1$ and $3$ have additive order $4$. Note $\Bbb Z_4$ is not a field with respect to addition and multiplication modulo $4$, since $2$ has no multiplicative inverse (it is a zero divisor).

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