Using the natural homomorphism $\mathbb Z$ to $\mathbb Z_5$ 
Prove that $x^4+10x^3+7$ is irreducible in $\mathbb Q[x]$ by using the natural homomorphism from $\mathbb Z$ to $\mathbb Z_5$.

So I would assume we should rewrite our polynomial, maybe as $(x + 10) x^3 + 7$? Then in terms of $\mathbb Z_5, (x+2\cdot 5)x^3+(5+2)\equiv (x+0)x^3+2=x^4+2$. Although I am not sure of this because I know that $F_5$ has no zero divisors, so maybe 10 cannot be canceled. Also, this didn't use the natural homomorphism, so this cannot be right. Could I have some help?

What you're missing about the "natural homomorphism" is:


*

*If your polynomial factors in $\mathbb{Q}[x]$, then it factors in $\mathbb{Z}[x]$.

*If your polynomial factors in $\mathbb{Z}[x]$, then that gives you a factorization in $\mathbb{F}_5[x]$.


So if it doesn't factor over $\mathbb{F}_5$, it doesn't factor over $\mathbb{Q}$.

There are a number of ways to show that $x^4 + 2$ is irreducible in $\mathbb{F}_5$. If $E$ is the splitting field, then $E / \mathbb{F}_5$ is a Kummer extension, since $\mathbb{F}_5$ has a fourth root of unity and you're adjoining a fourth root of an element of $\mathbb{F}_5$.
Even if you don't know about Kummer extensions, you can infer that $E$ has a sixteenth root of unity.
In any case, you can then deduce that $E = \mathbb{F}_{5^4}$. And there are a variety of ways to use that fact to show $x^4 + 2$ irreducible.
Exercise: To be sure you understand the overall argument, the same approach (with a little bit more work at the end) can be used to determine the degrees of the factors of $x^4 + 1$, and even what the factors are.
