Show that $f(x)=2x+1$ has a multiplicative inverse in $\mathbb{Z}_4[x]$, the integers mod 4. Show that $f(x)=2x+1$ has a multiplicative inverse in $\mathbb{Z}_4[x]$, the integers mod 4.
I don't really know where to start (besides dividing $1$ by $f(x)$). I thought having multiplicative inverses was one of the requirements for something to be a field.
Also, if the inverse is just $\frac{1}{2x+1}$, where do I go from there? How do I show that the inverse is in $\mathbb{Z_4}[x]$.
 A: $(2x+1)(-2x+1)=1-4x^2=1$ as polynomials in $\mathbb{Z}_4[x]$.
You can check Atiyah-MacDonald (chap.1- ex. 2.i) for a characterizazion of invertible elements in the ring $A[x]$ where $A$ is a generic commutative ring.
A: $$(ax+b)(2x+1)=2ax^2+x(2b+a)+b$$
$$\implies b=1+4\mathbb Z,a=2+4\mathbb Z$$
$$\implies (2x+1)(2x+1)=4x^2+4x+1=1\mod 4$$
The $ax+b$ is chosen to match the degree of the polynomial whose inverse is sought, although there may be cases where a smaller-degree polynomial is sufficient.
Follow-up question: is there a multiplicative inverse in $\mathbb Z_4[x]$ of $x-2$?
A: Every ring contains invertible elements, the point of fields is that EVERYTHING is invertible (except $0$). For example the ring $\mathbb{Z}$ is not a field yet it has two invertible elements $\pm 1$.
So anyway, for the question what do we have to do? We have to show that there is a polynomial $g(x)\in\mathbb{Z}_4[x]$ such that $f(x)g(x) = 1$. 
It is not enough to just write $\frac{1}{f(x)}$, that is not necessarily a polynomial!
Anyway, it is easy to see that $g(x)$ cannot be constant (just try all possibilities mod $4$). Next guess would be that a linear polynomial will do.
So suppose $g(x) = ax+b$. We want $(2x+1)(ax+b) = 1$.
Expanding gives:
$(2a)x^2 + (2b+a)x + b = 1$
Equating coefficients gives $b \equiv 1 \bmod 4$ and $a \equiv 2 \bmod 4$.
So $g(x) = 2x+1 = f(x)$, i.e. $f(x)$ is self inverse!
