I know that all finite fields of the same size are isomorphic to one another. I also know that if a polynomial $f(x)$ is irreducible over $\mathbb{Z}[x]$ and of degree $n$ then

$$ \frac{\mathbb{Z}_k[x]}{(f(x))} \cong \mathbb{Z}_{k^n}$$

For instance, we should have that $$ \frac{\mathbb{Z}_5[x]}{(x^2+2)} \cong \mathbb{Z}_{25}$$

However, I'm unsure of how to construct an explicit isomorphism between the two. I want an isomorphism so I can identify things like every generator of the field on the LHS (which will correspond to the inverse under the isomorphism of $2, 3, 4, 6, 7$ etc.). What is an isomorphism for this specific case and how do I then generalise this?

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    $\begingroup$ $\mathbb Z_{25}$ is not a field $\endgroup$ – J. W. Tanner Apr 30 '19 at 13:28
  • $\begingroup$ Okay, should it be $\mathbb{Z}_k^n$ (so in the specific example $\mathbb{Z}_5^2$)? $\endgroup$ – Adam Apr 30 '19 at 13:32
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    $\begingroup$ Keep in mind: a field has no non-zero zero divisors $\endgroup$ – J. W. Tanner Apr 30 '19 at 13:49
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    $\begingroup$ Perhaps you mean $\mathbb F_{25}$ instead of $\mathbb Z_{25}$ $\endgroup$ – J. W. Tanner Apr 30 '19 at 16:37

Your method of constructing a field of $25$ elements by the quotient ring of a quadratic polynomial is fine,

but it is a misconception that the ring of integers modulo $25$ is a field.

The ring $\mathbb Z_{25}$ has non-zero zero divisors -- for example, in it $5\times5=0$ even though $5\ne0$ --

so it is not a field. Integers modulo $n$ are a field when $n$ is prime.

The elements of the field of $25$ elements are of the form $a+b\alpha,$

where $a,b\in\mathbb F_5=\mathbb Z_5$ and $\alpha$ is a root of a quadratic polynomial that is irreducible in $\mathbb Z_5$.

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  • $\begingroup$ In other words, $\mathbb Z_{25}$ is not isomorphic to $\mathbb F_{25}$ $\endgroup$ – J. W. Tanner Apr 30 '19 at 19:29

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