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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 $Z_7[x]$.

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    I'm confused: if you meant $\;\Bbb Z_7=\Bbb Z/7\Bbb Z\cong\Bbb F_7=$ the field with seven elements, then a polynomial of degree $\;\ge1\;$ in $\;\Bbb F_7[x]\;$ cannot be associated with a polynomial of degree $\;0\;$ , so what do you really mean? Perhaps the unique monic polynomial is...$\;1\;$ ? – 2017-01-15
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    I've added the whole question to the main body of the body of the post, hope that clears things up. – 2017-01-15
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    Please do write properly all the mathematics in your question. As it is now it is almost impossible to understand... – 2017-01-15
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    The leading coef of $\,4 = 4 x^0\, $ is $\,4\,$. Of course scaling by its inverse yields a monic constant polynomial $\, 1 = 1 x^0\ $ (the only monic constant polynomial). – 2017-01-15
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    Ah, I see that was pretty obvious. I failed to think of it with an $x^0$ attached, which threw off me off. Thanks! – 2017-01-15

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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:

$$(5x^2+3)\cdot3=x^2+2\;,\;\;\text{and}\;\;3\;\;\text{is a unit, again}$$

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    I understand now, the lack of an $x$ confused me yet I failed to consider it as $4x^0$. I feel pretty silly now. Thanks for your help. – 2017-01-15