We need to construct a field of 25 elements.

By using a result : For a prime p and a monic irreducible polynomial $p(x)$ in $\Bbb F_p [x]$ of degree $n$ , then the ring $(\Bbb F_p [x] / <p(x)>)$ is a field of order $p^{n}$.

So , for the question problem above , we need a irreducible polynomial of degree $2$ in $\Bbb Z_5 [x]$.

Solution says , $x^{2} + x +1$ does our job , since it's irreducible in $\Bbb Z_5 [x]$.

My question : Can't we take $ x^{2} + 2$ , it has no zeroes in $\Bbb Z_5$ , and is irreducible. In every solution I've gone through , the polynomial $x^{2} + x+1$ is taken. Can't we consider $ x^{2} + 2$ ?

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    $\begingroup$ Yes, of course you can also take $x^2+2$. $\endgroup$ – egreg Feb 21 '16 at 13:42
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    $\begingroup$ Spoiler alert: It turns out that regardless of which irreducible polynomial of degree $n$ you use, you will get isomorphic fields of order $p^n$. $\endgroup$ – Ravi Feb 21 '16 at 13:45
  • $\begingroup$ By using $x^2+x+1$, the solutions you’ve seen have been adjoining a primitive cube root of unity to $\Bbb F_5$. Frankly, I find your method more natural and better suited to hand computation. $\endgroup$ – Lubin Nov 14 '17 at 0:25

Yes, you can take $x^2+2$: any degree two irreducible polynomial would do.

If you take $K=\mathbb{F}_p[x]/\langle x^2+2\rangle$ and denote by $u$ an element such that $u^2=3$, you can find in it a root of $x^2+x+1$: $$ (a+bu)^2+(a+bu)+1=0 $$ is the same as $$ a^2+3b^2+a+1=0,\qquad 2ab+b=0 $$ so we must take $a=-1/2=2$ and so $3b^2=3$; thus $2+u$ and $2-u$ are the solutions.


there is a theorem that says that if you have a field F and a polynomial f (x) of degree 2 or 3 in F [x] that does not have zeros in F, then f (x) is irreducible in F, therefore x ^ 2 + 2 is irreducible in Z5


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