Showing a factor ring is a field if it is an integral domain 
Let $K$ be a field, $0 \neq I \subsetneq K[x]$ be an ideal. Show that $K[x] \ / \ I$ integral domain $\Rightarrow$ $K[x] \ / \ I$ field applies.

My idea was to show that $K[x] \ / \ I$ is a finite integral domain, which would imply it is a field, as every element $\neq 0$ is invertible. So I need to show two things:


*

*$K[x] \ / \ I$ is an integral domain

*$K[x] \ / \ I$ is finite


In order to show it is an integral domain, I need to show that there are no zero divisors in $K[x] \ / \ I$. I'm stuck here, can you please help me to show that there are no zero divisors?
We know that $I$ is principle ($K[x]$ is a PID), and therefore generated by a single element. Can I use this to show $K[x] \ / \ I$ is finite?
 A: First, you certainly cannot show that $K[X]/I$ is finite since this is plainly not true when $K$ is infinite (as pointed out in  a comment).
Second, you do not have to show that $K[X]/I$ is an integral domain as the result is "if integral then field", again it is plainly not true that $K[X]/I$ is generally an integral domain. 
On the positive said: yes, you could show that $K[X]/I$  is finite (assuming that $K$ is finite) in the way your propose by showing that the elements of $K[X]/(f)$ essentially can be considered as polynomials of degree less than $\deg f$, via considering euclidean division.
Going back to your problem here is a proposal how to proceed (as you said you can restrict to principal ideal as every ideal is principal, let us denote $I = (f)$ in particular $f \neq 0$): 


*

*Show that $K[X]/(f)$ is an integral domain if an only if $f$ is an irreducible polynomial. 

*Conclude that $(f)$ is a maximal ideal, which implies $K[X]/(f)$ is a field. 
If the second point is not clear as you do not know the notion and or the result, use Bézout's identity to show that every polynomial $g$ that is not a multiple of $f$, for irreducible $f$, is invertible modulo $(f)$ as you can solve $P g + Q f = \gcd (f,g) = 1 $ in polynomials.
A: A direct proof: if $I\neq 0$, $I$ is generated by a monic polynomial of positive degree $d$ since $I$ is a proper ideal. As a consequence, the ring $A=K[x]/(f)$ is a $K$-vector space of dimension $d$; if we denote by $\xi$  the class $x+(f)$, a basis is the set $\{1, \xi, \dots, \xi^{d-1}\}$.
Now let's consider a non-zero element $u$  of $A$. If $A$ is an integral domain, multiplication by $u$ is an injective endomorphism of $A$. For an endomorphism of a finite dimensional vector space, injectivity is equivalent to surjectivity, hence $1$ is attained, i.e. there exists an element $v$ such that $uv=1$ – in other words every non-zero element has an inverse.
