The irreducibility of $t^4 + 2$ over $\mathbb{Z}_5$

Question

I'm reading Stewart's Galois Theory Third Edition Example 3.23, where we are proving that $f(t) = t^4 + 15t^3 + 7$ over $\mathbb{Z}$ is irreducible by showing its mapping to $\mathbb{Z}_5$ is irreducible.

First, the polynomial gets mapped to $t^4 + 2$ over $\mathbb{Z}_5$. Then Stewart says if it is reducible over $\mathbb{Z}_5$, then either it has a factor of degree 1, or it is a product of two factors of degree 2 (understandable). Then he claims "The first possibility gives rise to an element $x \in \mathbb{Z}_5$ such that $x^4 + 2 =0$" -- here I'm not following how this gives rise to such an equation for which we could check whether the factor of degree 1 is possible...Many thanks for help!

If there is a linear factor $t - a$ so that $t^4 + 2 = (t - a)p(t)$, then
$$a^4 + 2 = (a - a) p(a) = 0$$

in $\mathbb{Z}_5$ — , Dec 31 '15 at 21:57
As discussed here this polynomial remains irreducible over the bigger field $\Bbb{F}_{125}$. The answers there also show the irreducibility over $\Bbb{Z}_5$ because the main claim of the other question follows from that. — Jyrki Lahtonen, Dec 31 '15 at 23:10

score 3 · Accepted Answer · answered Dec 31 '15 at 21:58

3

(copied from my comment)

If there is a linear factor $t - a$ so that $t^4 + 2 = (t - a)p(t)$, then

$$a^4 + 2 = (a - a) p(a) = 0$$

in $\mathbb{Z}_5$.

answered Dec 31 '15 at 21:58

How do we know that $a$ must be in $\mathbb{Z}_5$? – nekodesu Dec 31 '15 at 22:01
3

@ShiyueLi Where else would it be? You're working over $\mathbb{Z}_5[t]$. – Alex Provost Dec 31 '15 at 22:05

The irreducibility of $t^4 + 2$ over $\mathbb{Z}_5$

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