irreducibility of a polynomial in several variables over ANY field The irreeducibility of a polynomial $f\!\in\!K[x_1,\ldots,x_n]$ in general depends on what the field $K$ is (for example, if $K=\mathbb{R}$, then $f=x_1^2+1$ is irreducible, but if $K=\mathbb{C}$, it is not). 
However, when $n\geq2$, there exist polynomials that are irreducible no matter what $K$ is (for example $x_i-x_j^d$ where $d\in\mathbb{N}$ is arbitrary and $i\neq j$, is always irreducible).
Question: Is there a command in any computer algebra system (preferably SINGULAR), that could (in some cases) confirm if a polynomial is irreducible over any field?
For example, in A Singular Introduction to Commutative Algebra, page 222, there is written:

Since there is mention of the quotient field $Q(A)$, this means $A$ is a domain, i.e. $x^4+6x^2y-y^3$ is irreducible.
Question: How can I show that $x^4+6x^2y-y^3$ is irreducible over any $K$?

This is a minor variation of Georges's answer. Assume by contradiction that $y^3-6x^2y-x^4$ is reducible in $K[x,y]$. Then we necessarily have 
$$
y^3-6x^2y-x^4=\Big(y-f(x)\Big)\Big(y^2+g(x)\, y+h(x)\Big)
$$
for some $f(x),g(x),h(x)$ in $K[x]$. This implies in particular $x^4=f(x)\,h(x)$. Thus $f(x)=a\, x^n$ with $0\le n\le4$ and $a\in K^\times$, and we easily rule out the five "possible" values of $n$.
(I upvoted the three other posts.)
EDIT. In order to prove that $y^3-6x^2y-x^4$ is irreducible in $K[x,y]$, it suffices to assume that $K$ is a domain. Indeed, in the above argument, we only need to know that (up to association) 
$$
x^4=x\cdot x\cdot x\cdot x
$$
is the only decomposition of $x^4$ as a product of irreducibles in $K[x]$.
