Given the extension $\frac{\mathbb Z_7[x]}{\langle 3+x^3\rangle}:\mathbb Z_7$ and $\alpha =x^2+\langle 3+x^3\rangle \in \frac{\mathbb Z_7[x]}{\langle 3+x^3\rangle}$, How can I compute the minimal polynomial $m_\alpha(x)$ and $ m_{\alpha^{-1}}(x)$ ?
Any tip? or Solution? will be appreciated.
So we have, in effect $\alpha=\beta^2$ where $\beta$ is a cube root of $-3$ modulo $7$. We need to find some linear dependence between $1$, $\alpha$, $\alpha^2$ and $\alpha^3$: precisely some $r$, $s$ and $t$ with $\alpha^3+r\alpha^2+s\alpha+t=0$. But $\alpha^2=\beta^4=-3\beta$ and $\alpha^3=-3\beta^2=2$ etc.
In general, when you have a field $K$ and an irreducible polynomial $p(x)$ (that we can assume is monic) on said field then $K[x]/\langle p(x)\rangle$ is essentially $K(\alpha)$ where $\alpha$ is a root of $p(x)$. If $\deg p(x) = n$ then this denotes an extension over $K$ od degree $n$, which means that as a $K$-vector space $K(\alpha)$ is generated by the basis $$(1,\alpha,\alpha^2,\ldots, \alpha^{n-1}).$$ This is due to the fact that if $$p(\alpha) = a_0 + a_1x + \ldots + a_{n-1}x^{n-1} + x^n$$ then $p(\alpha) = 0$ and therefore $$\alpha^n = -a_0 -a_1\alpha - \ldots - a_{n-1}\alpha^{n-1}.$$
If we have another element $\beta\in K(\alpha)$ then we can write $\beta$ as a linear combination of the elements of the $K$-base:
$$\beta = b_0 + b_1\alpha + \ldots + b_{n-1}\alpha^{n-1}$$
As far as I know there is no algorithm that quickly yields the minimal polynomial of $\beta$ but there is a simple way to find a candidate.
If $b_{k+1} = b_{k+2} = \ldots = b_{n-1} = 0$ then $\beta = b_0 + b_1\alpha + \ldots + b_k\alpha^k$. Then try to write $\beta^m$ in terms of $1,\beta,\ldots,\beta^{m-1}$ where $m = \dfrac{n}{\gcd(k,n)}$. This will ensure a polynomial that was $\beta$ as a root. Once you have it try to prove that it's irreducible or try to find a factor that has $\beta$ as a root.
If you know $m_\alpha(x)$ then computing $m_{\alpha^{-1}}(x)$ is really easy.
If $m_\alpha(x)$ is the minumal polynomial of $\alpha$ over the field $K$ then $K(\alpha) = K(\alpha^{-1})$ so the minimal polynomial of $\alpha^{-1}$ over $K$ has the same degree as that of $\alpha$.
If $$m_\alpha(x) = a_0 + a_1x + \ldots +a_{n-1}x^{n-1} + a_nx^n$$ then $$m_{\alpha^{-1}}(x) = a_0x^n + a_1x^{n-1} + \ldots + a_{n-1}x + a_n.$$
Indeed $$\alpha^n(a_0\alpha^{-n} + a_1\alpha^{-(n-1)} + \ldots + a_{n-1}\alpha^{-1} + a_n) = a_0 + a_1\alpha + \ldots + a_{n-1}\alpha^{n-1} + a_n\alpha^n = m_\alpha(\alpha) = 0$$
but $\alpha^n \neq 0$ so it must be that $a_0\alpha^{-n} + a_1\alpha^{-(n-1)} + \ldots + a_{n-1}\alpha^{-1} + a_n = 0$. That means that $\alpha^{-1}$ is a root of $a_0x^n + a_1x^{n-1} + \ldots + a_{n-1}x + a_n$ and since it has the same degree as $m_\alpha(x)$ it must be minimal.
