Proof that $K[X, Y]/(f)$ is an integral ring if $f$ is irreducible Let $K$ be a field and consider the polynomial ring $K[X, Y]$. If $f\in K[X, Y]$ is irreducible, then $K[X, Y]/(f)$ is an integral ring. 
How do you prove this? I know how to do this in the case of one variable but I have never treated such quotients with several variables.
Thank you.
EDIT
I just finished the proof, thank you all for your help !  
 A: The ring $K[X,Y]$ is known to be a unique factorization domain, so every irreducible element is also prime (you find the proof in any algebra textbook). Hence an irreducible polynomial generates a prime ideal.
A: Hint $\ K[X] $ is a UFD since it has a Euclidean division algorithm. Generally UFD $D\Rightarrow$ UFD $D[Y].\,$ Combining we deduce $\,R = K[X,Y] \cong (K[X])[Y]$ is a UFD, therefore the irreducible $f$ is prime, hence $\, R/f\,$ is a domain.
Why is UFD $D\Rightarrow$ UFD $D[Y]$ true? The usual proof uses Gauss's Lemma. However that obscures the essence of the matter. Suppose we have a domain where every nonzero nonunit factors into irreducibles. To show it is a UFD requires showing that every irreducible is prime. Suppose we already know that some subset $S$ of irreducibles are prime. Then we can simplify matters by "ignoring" elements of $S$ in factorizations. This we achieve by forcing them to be units, i.e. by working in the subring of its quotient field obtained by adjoing $1/s$ for all $\,s\in S.\,$ It is quite easy to show that this enlarged ring is a UFD $\iff$ the original ring is a UFD.
Now we apply this to $\,D[Y],\,$ with $S$ being all primes of the UFD  $D$. Adjoining their inverses yields $F[Y]$ where $F$ is the fraction field of $D$. But this is a UFD, having a Euclidean division algorithm. Therefore, by the prior paragraph, we conclude that $D[Y]$ is a UFD too.
This is sometimes called "Nagata's trick", but I suspect that the basic idea is much older. The method of adjoining inverses to kill uninteresting objects is a special case of  localization.
A: Hints:
++ If $\;R\;$ is a commutative unitary ring, then the ideal $\;I\le R\;$ is prime iff the quotient ring $\;R/I\;$ is an integral domain.
++ A principal ideal $\;\langle f\rangle\le K[X,Y]\;$ is prime iff $\;f\in K[X,Y]\;$ is irreducible
