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I assumed that if we quotient $Z$ by the ideal generated by 4, it will be the same as the field $F4$. It turns out not to be the case because $Z/(4)$ is not a field.

In general, when is $Z/(n) = Fn$?

Lana
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    We have $\Bbb Z/n \Bbb Z \cong \Bbb F_n$ as rings only when $n$ is prime. By the way, the notation $\mathbb F_n$ is used only when $n$ is a prime power. – Watson Apr 01 '17 at 18:35
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    They are not even isomorphic as abelian groups, since $\Bbb F_4$ has characteristic $2$ you can see $2x=0$ for all $x\in\Bbb F_4$. But in $\Bbb Z/(4)$ you see $2[1]=[2]\ne [0]$. – Adam Hughes Apr 01 '17 at 18:39
  • I thought the elements of F4 are 0, 1, 2, 3. 2x1 = 2 in F2, no? – Lana Apr 02 '17 at 17:55
  • We have $x+x=0$ and $2\cdot 1=2$ in F4 but in F2 there is no 2 in $2\cdot 1$ because $2\equiv 0(mod 2)$ – QuantumPotatoïd Mar 13 '21 at 11:10

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