Let $\mathbb{K}$ be a field. Given an ideal $I \subset \mathbb{K}[x_1,\dots, x_n]$ and a monomial order we can consider the ideal $LT(I) = (lt(f) \ | \ f\in I )$, where $lt(f)$ denotes the leading term of the polynomial $f$. It can be shown that if $LT(I)$ is prime then $I$ is prime, but the viceversa does not hold (e.g. $(x^2 + 1)\subset \mathbb{Q}[x]$). I wonder then what kind of relation there is between the irreducibility of $LT(I)$ and the irreducibility of $I$.
Formally my question is:
$LT(I)$ irreducible $\Rightarrow$ $I$ irreducible?
$I$ irreducible $\Rightarrow$ $LT(I)$ irreducible?
In one variable, $LT(I)$ will always be irreducible, since $I=(x^n)$ is irreducible. Hence the answer to your first question is negative, i.e. take $I=(x^2-1) = (x-1) \cap (x+1)$. This is not irreducible, but $LT(I)=(x^2)$ is.
The second question can also be answered negatively: $I=(xy-1) \subset K[x,y]$ is irreducible, because it is prime. $LT(I)=(xy)$ is not.