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Give an example of a natural number $n > 1$ and a polynomial $f(x) ∈ \Bbb Z_n[x]$ of degree $> 0$ that is a unit in $\Bbb Z_n[x]$.

For this is set $n=2$. So then $f(x) = x \in \Bbb Z_2[x] $. This is a unit because $f(1)f^{-1}(1) = 1$. I don't think this is correct as I can't use $f(0)$.

Find all the polynomials of degree less or equal to $2$ in $\Bbb Z_2[x]$.

I don't really know how to do this. I thought that since we are working in $\Bbb Z_2$,I can't really have a polynomial of degree $2$. This is confusing me But if I have to: $ax^2, ax^2 + bx, ax^2 + bx + c, ax^2 + c, bx, bx + c, c: a,b,c \in \Bbb Z_2$

TfwBear
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  • If the question had not said less than or equal to $2$, is the maximal degree the polynomial can have $1$? @SeanI – TfwBear Apr 28 '16 at 21:59
  • degree of the polynomial is irrelevant to the characteristic of the ring. It will be relevant if you evaluate at a point. I think this is what confused you. Your list seems to be correct. But I think the question wants you to explicitly list the polynomials. For the first part this might help http://math.stackexchange.com/questions/19132/characterizing-units-in-polynomial-rings – ugur efem Apr 28 '16 at 22:06

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No, $x$ is not a unit; there is no polynomial $p(x)$ such that $xp(x)=1$. This is because $\deg xp(x)=1+\deg p(x)\ge 1$.

Actually, one shows that, for a polynomial of positive degree to be a units, it is necessary that its leading coefficient to be a zero-divisor, which implies $n$ to be composite. A simple example: in $\mathbf Z/4\mathbf Z[x]$, we have: $$(2x+1)^2=4x^2+4x+1=1.$$

As to the polynomials of degree at most $2$ in $\mathbf Z/2\mathbf Z[x]$ you just have to write out all possible sets of coefficients: $$\{0,1, x,1+x, x^2, 1+x^2, x+x^2, 1+x+x^2\}.$$

Bernard
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  • I gave one such example. – Bernard Apr 28 '16 at 22:23
  • Sorry. I just read back through that. So how is $2x+1$ a zero divisor? What is its inverse in $\Bbb Z_4$? And why does it imply that n has to be composite? – TfwBear Apr 28 '16 at 22:26
  • $2x+1$ is not a zero-divisor. It can't be, since it is a unit, and its inverse is itself. What I said is the leading coefficient has to be a zero-divisor. Indeed, mod$4$, $2$ is nilpotent. – Bernard Apr 28 '16 at 22:30
  • Ok. That makes sense. So why would the leading coefficient have to be a zero divisor? Is there some theorem that states that? – TfwBear Apr 28 '16 at 22:36
  • Because if it isn't, the degree of the product will be the sum of the degrees of the factors, hence it can't be $0$, the degree of the constant $1$. – Bernard Apr 28 '16 at 22:38
  • You're welcome! I'll remove my comments about this point. They have no interest for someone else. – Bernard Apr 28 '16 at 22:41