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Let's say I have an irreducible polynomial $f(x) \in\mathbb Z/(2)[x]$ with a degree n that is at least 2 or greater.

How would I go about proving that $\mathbb Z/(2)[x]/(f(x))$ is a finite field of size $2^n$? All that I've gotten so far is why there will be $2^n$ elements in the field, but not how they form a field.

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In general, if $R$ is a ring and $I$ is an ideal of $R$ then the quotient $R/I$ is a field if and only if the ideal $I$ is maximal -- by maximal, I mean that there is no other ideal $J$ with $I\subsetneq J\subsetneq R$. In your situation, why is it true that the ideal generated by an irreducible polynomial is maximal?

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$\mathbb{Z_2}[x]$ is a PID and $(f(x))$ is a maximal ideal- so quotient is a field.

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Being a polynomial ring over a field, $\,\Bbb F_2[x]\,$ has a Euclidean algorithm, so every ideal is principal $\,I = (f)\,$ for $\,f =$ gcd of all elements of $I.\,$ For principal ideals we have $ $ contains = divides, $ $ i.e. $(g)\supset (f)\iff g\mid f,\,$ so $\,(f)\,$ is maximal $\iff$ $\,(f)\,$ has no proper container $\,(g)\,$ $\iff$ $\,f\,$ has no proper divisor $\,g\,$ $\iff\,f\,$ is irreducible. Finally, generally $\,R/I\,$ is a field $\iff I\,$ is maximal.

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