This is from Ian Stewart's book on Galois theory, I am looking at irreducible polynomials.
It talks of irreducibility over mod. It takes as an example, $f(t)=t^4+15t^3+7$ over integers, and asks us to consider it over $\mathbb{Z}_5$ which makes it $t^4+2$. If this is reducible over $\mathbb{Z}_5$, then it either has a factor of degree 1, or a product of 2 factors of degree 2. This is fine, I understand.
What I don't understand is, "The first possibility gives rise to an element $x \in \mathbb{Z}_5$, such that $x^4+2=0$"
Simply, why? Why does "having a factor of degree 1" imply this? This first possibility is essentially, $(t+a)(t^3+bt^2+ct+d)=t^4+2$ in $\mathbb{Z}_5$, yes? I tried expanding this and sought if any of the coefficients comparison leads to me concluding that there must be an element of $x^4+2=0$ in the field but no.
I don't see why this conclusion is obtained. Can someone explain to me in detail please?
Recall that the polynomial having a linear factor $t+a$ over $\mathbb{Z}_5$ implies, indeed is equivalent to, the polynomial having a root at $-a \in \mathbb{Z}_5$. This is not specific to this polynomial but is general fact for polynomials over fields.
But to see it explicitly in your explicit factorization if you plug in $-a$ you get $0$ on the left as you have a factor of $(-a)+ a= 0$ and thus you have $0$ on the right so $(-a)^4 + 2 = 0$, so $-a$ is the $x$ you seek.
Thus if $t^4 + 2$ had a linear factor, there would be a root $x \in \mathbb{Z}_5$, that is some $x$ such that $x^4 + 2 = 0$.
This is however impossible, as $x^4$ can only be $1$ or $0$ for $x \in \mathbb{Z}_5$. (This follows from little Fermat theorem, or by just checking all cases.)
Thus one can exclude the existence of a linear factor, and can focus on the only remaining case that $t^4 + 2$ is the product of two quadratic polynomials.