My understanding is that to get a unique monic polynomial you need to take the leading coefficient and multiply it by its inverse.
Using $5x^3+3$ in $Z_7$ as an example,
The inverse of 5, the leading coefficient, is 3.
$3(5x^3+3)=15x^3+9$,
which is $x^3+2$ (mod 7)
However, I don't know how to apply this to just 4. If I imagine it as $0x+4$ or something zero doesn't have an inverse. I'm quite confused and would appreciate some help!
The actual question: (a) Let f(x) be a polynomial in Q[x] with $f(x) = a_0 + a_1x + · · · + a_(n−1)x^(n−1) + a_nx^n$.
It says the summation notation but I don't know how to format that.
Write down each of the following products using the summation notation, taking care in your answer to ensure that the summands each contain only a single power of x.
(i) $xf(x)$
(ii) $(x + 1)f(x^4)$
(iii) $(b_0 + b_1x + b_2x^2)f(x)$ [5]
(b) Verify that $Z7[x]$ satisfies the ring axiom R6.
There's another summation that I can't do.
(c) (i) Write down the units in $Z_7[x]$. [2]
(ii) Find the unique monic polynomial that is an associate of $4$ in $Z_7[x]$. [2]
(iii) Find the unique monic polynomial that is an associate of $5x^2 + 3$ in $Z_7[x]$.
Then yes: the wanted monic polynomial is $\;1\;$ , since $\;4\cdot2=1\;$ and $\;2\;$ is a unit in $\;\Bbb F_7[x]\;$ .
For the other one:
$$(5x^2+3)\cdot3=x^2+2\;,\;\;\text{and}\;\;3\;\;\text{is a unit, again}$$