Find a nontrivial unit polynomial in $\mathbb Z_4[x]$ Find a unit $p(x)$ in $\mathbb{Z}_4[x]$ such that $\deg p(x)>1$
What I know:
A unit has an inverse that when the unit is multiplied by the inverse we get the identity element.
But I am confused by the concept of degree
"If $n$ is the largest nonnegative number
for which an $a_n \neq 0$, we say that the degree of $f$ is $n$."
So I am looking for a polynomial in $\mathbb{Z}_4[x]$ that has an inverse..
Also I am not sure exactly what is $\mathbb{Z}_4[x]$?
Any ideas on how to get started on this problem?
 A: Take $p(x) = 2x^2+1$. Observe that $p(x)^2 = (2x^2+1)^2 = 4x^4+4x^2+1 = 1$, and $p(x)$ has degree $2$.
A: The key is that $\mathbb{Z}_4$ (the ring of residue classes modulo $4$) has nilpotent elements.
Suppose $A$ is a commutative ring and that $a$ is a nilpotent element; then $1-a$ is a unit in $A$, because
$$
1-a^n=(1-a)(1+a+a^2+\dots+a^{n-1})
$$
for all $n>1$, and we can take $n$ such that $a^n=0$.
Since $[2]$ (the residue class of $[2]$ modulo $4$) is nilpotent, also $[2]x$ is and therefore $[1]-[2]x$ is invertible. Indeed
$$
([1]-[2]x)([1]+[2]x)=[1]-[2]^2x^2=[1]
$$
because $[2]^2=[4]=[0]$.
I leave to you finding a degree $2$ unit.

 Starting from the facts that the sum of nilpotent elements is nilpotent and that $ax^r$ is nilpotent if and only if $a$ is nilpotent, we can prove that a polynomial $a_0+a_1x+\dots+a_kx^k$ is a unit in $A[x]$ if and only if $a_0$ is a unit in $a$ and $a_1,\dots,a_k$ are nilpotent.

A: $\Bbb Z_{4}[x]$ is, by definition, the set $\{ a_{0} + a_{1}X + a_{2} X^{2} + a_{3}X^{3} + \dots + a_{n}X^{n} | a_{i} \in \mathbb{Z}_{4}, n \in \Bbb N \}$.  In other words, it is the ring of polynomials with coefficients in $\Bbb Z_{4}$.
Now, we want to find a polynomial that has a multiplicative inverse with the degree of the polynomial at least 2 (i.e., there should be a non-zero $x^{n}$ term for some $n \geq 2$).
Consider $1 - 2x^{3}$ and $1 + 2x^{3}$.  Multiplying these two together gives the form $(a - b)(a + b)$ which is the factored form of $a^{2} - b^{2}$.
But $(1)^{2} - (2x^{3})^{2} = 1 - 4x^{6} = 1 - 0 = 1$.  Thus, $p(x) = 1 - 2x^{3}$ is a polynomial of degree $> 1$ which is a unit (i.e., is invertible).
So, take any $n \geq 2$ and by the above argument it follows that $1 - 2x^{n}$ is a unit since multiplied against $1 + 2x^{n}$ gives $1$ and the degree is $n$ which is $\geq 2$.
