Is there an algorithm to decide if an ideal contains monomials? Let $I\subset k[x_1,\dots,x_n]$ be an ideal in a polynomial ring in commuting variables. Is there a procedure to decide if $I$ contains a monomial and possibly to find one? 
Gröbner bases come to mind, but for example, $$I = \langle x-y-z, y^4z^2+2y^3z^3+y^2z^4  \rangle = \langle  x-y-z,x^2y^2z^2\rangle$$ has $$\left\{x^4y^2-2x^3y^3+x^2y^4, y^4z^2+2y^3z^3+y^2z^4, x^4z^2-2x^3z^3+x^2z^4, x-y-z\right\} $$ as its universal Gröbner basis, which makes me pessimistic about Gröbner methods.  In this example, the symmetry can be broken by computing a primary decomposition which reveals the monomials, but what can be said in general?
 A: Computing colon ideals is pretty quick. You could colon out the
variables in order. If the ideal changes, record the variable that
worked, and go back to the beginning of the list.
Either you get to the unit ideal, in which case you found 
the lex-first monomial that's in the ideal, or you make it to the
end of the list, in which case there's no monomial in your ideal.
EDITS:


*

*There's no need to loop back to the beginning of the list. Do $x_1$ to completion, then $x_2$, and so on.

*Doing $x_i$ only requires a Gröbner basis for an elimination order for $x_i$, and the colon ideal will come with a Gröbner basis again, for free. So you don't even need a universal Gröbner basis, just $n$ elimination Gröbner bases.

*You could also do this by coloning out $x_1 x_2 \cdots x_n$, which is very close to 
François Brunault's answer comment. But my understanding is that colon ideals are computed using the elimination of a new variable, so I doubt this is actually faster than what I'm suggesting.

*You don't even need those full Gröbner bases, just ones that are Gröbner enough; see e.g. theorem 6.3 of http://arxiv.org/abs/dg-ga/9706003 where we use a criterion like this to compute a cohomology ring.
ANOTHER:
You could slice with random hyperplanes, and whenever you get isolated points, see if each of those points has some coordinate $=0$.
