How does extending the polynomial field F[x] work if it has multiplicative inverses? Ok I will try and make this as clear as possible.
I'm pretty new to Ring theory and I just finished watching this video:https://www.youtube.com/watch?v=1W87jB0I0P0&index=26&list=PLAvgI3H-gclZ-DYOVyyTkJBUXZ_R1JEZP
It talks about extending a field F[x] by finding the quotient ring using a principal ideal (p(x)) my understanding is that this adds the solution to the polynomial p(x) = 0 to the field thus extending it. What I don't understand Is how you can generate this principal ideal since you will multiply p(x) by its multiplicative inverse adding the multiplicative identity to the ideal and hence giving the entire ring as the ideal no? Then obviously only one coset exists which means that if anything you have shrunk the ring not extended it. Where am I going wrong? 
 A: When you say "extending a field $F[x]$", I think you real mean "a field $F$".  The polynomial ring $F[x]$ is not a field, but you can use it to form extension fields of $F$.  
Your confusion stems partly from you using $(p(x))$ to mean two different things. (It's not your fault -- the notation is made as simple as possible, but the cost is this ambiguity.)  Whenever you write a principal ideal $(p(x))$, it must be understood what ring you are working in.  For instance, the ideal $(2)$ makes sense both in the ring of integers $\mathbb{Z}$ and in the larger ring $\mathbb{Q}$.  In the latter, the ideal is the whole ring, since $2$ has a multiplicative inverse in $\mathbb{Q}$.  In $\mathbb{Z}$, $(2)$ consists of only the even integers, so is not the whole ring.
Finally, the ideal $(p(x))$ in $F[x]$ should also be kept distinct from the principal ideal $(\overline{p(x)})$ in a quotient ring of $F[x]$.  When $(p(x))$ is maximal and so the quotient ring $F[x]/(p(x))$ is a field, the element $\overline{p(x)}$ is $0$, has no multiplicative inverse, so the principal ideal $(\overline{p(x)})$ is the zero ideal, not the whole field.
