Q:

Zero set of homogeneous irreducible polynomial over reals

To show that the zero set of an irreducible polynomial in $k[X_1,\dots,X_n]$ is irreducible in $\mathbb{A}^n_k$, we need to assume that the field $k$ is algebraically closed.  Indeed, for $f(x,y)=(x^2-1)^2+y^2$, we have that $f$ is irreducible over $\mathbb{R}$ but the zero set is not irreducible (in fact it is disconnected). 
But, what if the polynomial is also homogeneous?  Is the zero set of an irreducible, homogeneous polynomial an irreducible algebraic set over arbitrary (not necessarily algebraically closed) fields?  I cannot find any counter-example to this. 
More specifically, my question is about the zero set of the determinant polynomial.  Consider, the polynomial ring $k[X_{11}, X_{12},\dots X_{nn}]$ and the polynomial $P=\det (X_{ij})_{n\times n}$. Then, $P$ is irreducible and homogeneous.  Is the zero set of $P$ irreducible over the reals? 

A:

For the main question, the answer is no, not necessarily.  

For example, let $f = x^2y^2 + z^4$.

Then $f$ is irreducible in $\mathbb{R}[x,y,z]$.

It's easily seen that 
$$V(f) = \{(x,y,z) \in \mathbb{R}^3 \mid z=0,\;\text{and at least one of}\;x,y\;\text{is zero}\}.$$

Consider the ideal $I = I(V(f))$

Then $xy\in I$, but $x \notin I$ and $y \notin I$, so $I$ is not a prime ideal of $\mathbb{R}[x,y,z]$.

Hence $V(f)$ is not irreducible.