Let $K$ be a field. Show that $x^2-yz$ is irreducible in $K[x,y,z]$. Deduce that $x^2-yz$ is prime.
If it is $K[x]$, then there are several methods which can be used to check whether a given polynomial is irreducible. But how do we check that when we have a polynomial of several variables? No idea how to do it. Any theorems? Moreover, in general, it is not true that irreducible elements are prime. So, how can I deduce the last result? Any help is appreciated.
For a polynomial in several variables one can sometimes apply Eisenstein's criterion. Your particular polynomial is Eisenstein at the prime $(y)$ in $(K[y,z])[x]$, for example, but also at $(z)$ in $(K[y,z])[x]$.
But looking at your polynomial in $(K[x,y])[z]$, it doesn't seem to be Eisenstein. After all if $x^2\in\mathfrak{p}$ for some prime $\mathfrak{p}\subset K[x,y]$, then also $x\in\mathfrak{p}$ and hence $x^2\in\mathfrak{p}^2$. Luckily there's a trick:
Your polynomial $P(x,y,z)=x^2-yz$ is irreducible if and only if $P(A(x,y,z))$ is irreducible, where $A$ is any invertible linear transformation over $K$, i.e. $A\in\operatorname{GL}_3(K)$. Choosing one wisely we look at $$P(x,y,z-x)=x^2-y(z-x)=-yz+x(x+y),$$ which is Eisenstein at both $(x)$ and $(x+y)$ in $(K[x,y])[z]$.
If it is reducible, you can write it as the product of two linear polynomials. For degree reasons it has to be $p(x,y)q(x,z)$. Now $$x^2-yz=(ax+by)(cx+dz)\iff ac=1,\ bd=-1,\ ad=0,\ bc=0.$$ These equations are incompatible.
In UFDs, irreducible elements are prime.
A quadratic form over a field $K$ of rank $\ge 3$ is irreducible. Since a product of two linear forms is of rank $\le 2$. ( assume $\operatorname{char}K \ne 2$ for comfort).
