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?
 A: I will do all arithmetic in $\mathbb{Z}_5$.  I will repeatedly use the fact that multiplication is commutative in this field.  Consider
$$
0=x^4+2
$$
We see that $x \ne 1$, and $x \ne 2$ since $2^4 = 1$.  Also, $x \ne 3$, since $3^4=1$.  Finally $x \ne 4 = -1$, since $-1^4=1$.  So there is no root to this polynomial in $\mathbb{Z}_5[x]$.  
Hence we cannot write
$$
x^4 + 2 = (a x + b)(cx^3+dx^2+ex+f)
$$
because $x = a^{-1}(-b)$ would be a root.
Now let's check if
$$
x^4+2 = (ax^2+ex+b)(cx^2+fx+d)
$$
makes sense.
First, the quartic term implies
$$
ac=1.
$$
Second, the constant term implies that
$$
bd=2.
$$
Third, the cubic term implies that
$$
af+ce = 0
$$
Fourth, the coefficient on $x$ implies that
$$
bf+de=0\\
$$
Writing these last two equations as a system, we have
$$
\left[
\begin{array}{cc}
c & a \\
d & b \\
\end{array}
\right]
\left[
\begin{array}{c}
e \\
f \\
\end{array}
\right] =
\left[
\begin{array}{c}
0 \\
0 \\
\end{array}
\right]
$$
I claim that matrix is not singular.  If it were, then its determinant would be zero.  Then we could write:
$$
cb-ad=0\\
cb=ad\\
acb=a^2d\\
b=a^2d \\
bd=a^2d^2 \\
2=(ad)^2 \\
$$ 
and $2$ is not the square of any number in $\mathbb{Z}_5$.  Hence
$$
e = f = 0
$$
Considering the coefficient on $x^2$, we see that
$$
ad+bc+ef=0\\
ad+bc=0\\
c^{-1}d+bc=0\\
d+bc^2=0\\
d=-bc^2
$$
Since $bd=2$, we may write
$$
2=bd=-b^2c^2=-(bc)^2\\
(bc)^2=-2=3
$$
There is no element of $\mathbb{Z}_5$ whose square is $3$.  
So the polynomial is not reducible over $\mathbb{Z}_5[x]$.
A: The other answers do not address why there cannot be a quadratic factor.  I don't know if there's a clever thing I'm missing, but here's one way to see.  Assume the quadratic splits over $\mathbb{Q}$ as 
$$
x^4 + 2 = (x^2+ax+b)(x^2+cx+d) 
$$
Then, passing to $\mathbb{F}_5$ we get the equations
$$
\begin{eqnarray}
d+ac+b = 0 \\
a+c = 0 \\
ad+bc=0 \\
bd = 2
\end{eqnarray}
$$
From which we get
$$
\begin{eqnarray}
a(d-b)=0
\end{eqnarray}
$$
So either $a=0$ or $b=d$.  If $b=d$ then $b^2 = 2$, but 
$$
\begin{eqnarray}
0^2=0\\
1^2=1\\
2^2=4\\
3^2=4\\
4^2=1
\end{eqnarray}
$$
so, $b^2=-2$ is not possible.  Therefore $a=0$.  But then from $d+ac+b=0$ we get $d=-b$, so $-b^2=2$, and $-2$ is not a square either.  Therefore we can't factor $x^4+2$ in to quadratic polynomials over $\mathbb{F}_5$.  Since a factorization over $\mathbb{Q}$ would descend to $\mathbb{F}_5$, there is also no factorization in to quadratics over $\mathbb{Q}$. 
A: 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.
