So I am given the following polynomial $f(x)=x^2+2x+3\in \mathbb{Z}_5[x]$ and asked to show that it is first irreducible and then that it is primitive.
So in terms of irreducibility, this is equivalent to saying that $f(x)$ has no roots in $\mathbb{Z}_5$, which is easy enough to check:
$f(0)=3\neq0,f(1)=1\neq0,f(2)=1\neq0,$
$f(3)=3\neq0,f(4)=2\neq0,f(5)=3\neq0$
Hence it is irreducible. Now my issue is showing that it is primitive.. I have seen some examples and attempted to follow them but feel as though I'm not really understanding the concepts.
We can deduce $|F^*|=5^2-1=24$ so, $o(\alpha)=24$. We know $24=3\times2^3$ and that $12,8$ are the proper maximal factors,so we just need to show that the following is true:
$\alpha^{12}\neq1$ and $\alpha^{8}\neq1$.
Now I'm not sure if any of this is right and I'm not sure, assuming it is correct, how to show the two equations above are actually true.
Any help is greatly appreciated!
This solution requires manipulation of our original equation $f(\alpha)=\alpha^2+2\alpha+3$.
First I will show that $\alpha^8 \neq 1$:
We can see directly $\alpha^2=3\alpha+2,$ and
$2\alpha=4\alpha^2+2\implies\alpha=\alpha^2+1$.
So, $\alpha^4=(3\alpha+2)^2=9\alpha^2+12\alpha+4=4\alpha^2+2\alpha+4$. We can sub in our $\alpha^2$ and simplify as follows: $4(3\alpha+2)+2\alpha+4=12\alpha+8+2\alpha+4=4\alpha+2$
Now, $\alpha^8=(\alpha^4)^2=(4\alpha+2)^2=16\alpha^2+16\alpha+4=\alpha^2+\alpha+4=3\alpha+2+\alpha+4=4\alpha+1$. Hence $\alpha^8 \neq1$.
Next, to show that $\alpha^{12}\neq1$:
$\alpha^{12}=\alpha^8\alpha^4=(4\alpha+2)(4\alpha+1)=16\alpha^2+12\alpha+2=\alpha^2+2\alpha+2=3\alpha+2+2\alpha+2=4$. Hence we have also shown $\alpha^{12}\neq1$.
Therefore, we can conclude $f(x)$ is primitive.