An irreducible polynomial of degree $4$ in $\mathbb{Z}_5[x]$ Q: Find an irreducible polynomial of degree four in $\mathbb{Z}_5[x]$.
My Answer: $x^{4} - 2$
Is my answer correct?

This is correct, but you need to show it's irreducible. Here's one way to do it:
We immediately verify that $x^4-2$ has no roots in $\mathbb Z _5$. Thus, if it is reducible, it must split into two monic quadratic terms. Say $x^4-2=(x^2+ax+b)(x^2+cx+d)=x^4+x^3(a+c)+x^2(b+d+ca)+x(ad+bc)+bd$. Now, matching coefficients, we must have that $bd=-2=3, a+c=0,b+d+ca=0, ad+bc=0$. First, consider $bd=3$. This then holds true for $b=3,d=1$ or $b=4,d=2$. Thus, we must have $b+d+ac=0$ implying that $ac=1$, in the first case, or $ac=4$, in the second. The values for the pair $(a,c)$ are then $(1,4), (2,3), (0,0), (3,2), (4,1)$. Note that there are two pairs each where $ac=1$ or $4$.  First, let $(b,d)=(3,1)$, so $(a,c)=(2,3)$ or $(3,2)$. Then $ad+bc=2+9=1$ or $ad+bc= 3+6=9$, neither of which are equal to $0$,  so neither of these work. Now, let $(b,d)=(4,2)$, so $(a,c)=(1,4)$ or $(4,1)$. Again, we see $ad+bc = 4+8 =2$ or $16+2=3$, so neither of these work either. Thus, no two quadratic polynomials multiply to $x^4-2$, so it is irreducible.
