What is the inverse of an element of polynomial ring over finite field?

Question

Let's consider the polynomial ring $\mathbb{F}_q[x]$. How to find the inverse of an element of this ring. For example, If I'm working over $\mathbb{Z}_7[x]$, what is the inverse of $x^2+x+1$. This is a reducible element. Does irreducibility effects to find the inverse. If so, what is the inverse of $x+1$. Please clarify this.

Answer 1

As pointed out by Arturo Magidin in his comment to the question, the only invertible elements of $\Bbb K[x]$, where $\mathbb{K}$ is any field, are the non-zero constants, the non-vanishing polynomials of degree $0$.

I think this is most easily seen via simple degree argument; if

$p(x), q(x) \in \Bbb K[x], \tag 1$

then

$\deg p(x)q(x) = \deg p(x) + \deg q(x), \tag 2$

as may be seen by writing

$p(x) = \displaystyle \sum_0^{\deg p} p_i x^i, \; p_i \in \Bbb K, \tag 3$

$q(x) = \displaystyle \sum_0^{\deg q} q_j x^j, \; q_j \in \Bbb K; \tag 4$

we then have

$p(x)q(x) = \left (\displaystyle \sum_0^{\deg p} p_i x^i \right ) \left (\displaystyle \sum_0^{\deg q} q_j x^j \right ) = \displaystyle \sum_{k = 0}^{\deg p + \deg q} \sum_{j = 0}^k p_j q_{k - j} x^k$ $= p_{\deg p}q_{\deg q} x^{\deg p + \deg q} + \displaystyle \sum_{k = 0}^{\deg p + \deg q - 1} \sum_{j = 0}^k p_j q_{k - j} x^k, \tag 5$

which shows the leading term of $p(x)q(x)$ is in fact $p_{\deg p}q_{\deg q}x^{\deg p + \deg q}$; (2) then follows from this observation.

Now suppose $p(x)$ and $q(x)$ are inverses of one another in $\Bbb K[x]$; then

$p(x)q(x) = 1, \tag 6$

whence, via (2),

$\deg p(x) + \deg q(x) = \deg p(x)q(x) = \deg 1 = 0; \tag 7$

since

$\deg p(x), \; \deg q(x) \ge 0, \tag 8$

(7) is only possible if

$\deg p(x) = \deg q(x) = 0, \tag 9$

that is,

$p(x) = p_0 \in \Bbb K, \; q(x) = q_0 \in \Bbb K. \tag{10}$