Properties of ideals preserved under extension of scalars The motivation for this question comes from a question in a book by a certain R.H dealing with geometrically reduced and irreducible schemes.
Let $k \subset K$ be algebraically closed fields and let $I$ be a prime ideal in the polynomial ring $k [x_1 \ldots x_n]$, generated, say, by $f_1, \ldots f_m$. Suppose that $I$ is prime or that the radical of $I$ is prime. These properties, I think, should be preserved if we consider $I$ as an ideal in the larger polynomial ring of $K$.
I have at least two arguments for that:
1) The first follows by translating those ideal theoretic to a system of polynomial equations and inequations in $K$ with coefficients in the small field $k$. By Hilbert's Nullstellensatz,  if there's a solution in the big field, then there's already a solution in the small one.
2) Use quantifier elimination. Every extension of algebraically closed fields is an elementary extension (in the sense of model theory). Translate the fact that I is not prime into a first order formula with parameters in the small field $k$. If there's a witness in the larger field, there is a witness in the small one already (Here one should be a little bit careful: each formula only handles polynomials of bounded degree. But compactness ensures that if an ideal is not prime, then there is a bound to the degree of the polynomials witnessing this fact as a function of the degrees of its generators).
However, I am not sure how to make these notions precise and write them down elegantly. I will appreciate help
 A: This is indeed true. As Keenan's answer stated, the geometric version of the notion you're talking about is geometric irreducibility and reducedness. 
Consider the scheme $X$ cut out by $I$. Then $X$ is irreducible if and only if $\sqrt{I}$ is prime. Then $X$ is geometrically irreducible if it stays irreducible after any base extension. Similarly, $X$ is reduced if and only if $\sqrt{I} = I$ and $X$ is geometrically reduced if it stays reduced after base extension. Your case of $I$ being prime is the case of being both geometrically reduced and geometrically irreducible. 
Then under this setup, indeed it is true that a scheme that is reduced (irreducible) over an algebraically closed field is geometrically reduced (geometrically irreducible). Therefore, in fact you can weaken your assumption of $K$ being algebraically closed. You just need $k$ to be algebraically closed. 
For proofs of these statements, see sections 4 and 6 of this chapter of the stacks project: http://stacks.math.columbia.edu/download/varieties.pdf. You'll have to go back to some earlier sections where they reference algebra results but you should be able to chase around a full proof. 
A: EDIT: The answer below ignores the fact that the OP wants to assume his fields are algebraically closed, something I missed when I first read the question. I'll leave this answer for now as an illustration of what goes wrong when the base field is not algebraically closed.
This isn't true. If it were then the notions of geometric reducedness and irreducibility would be equivalent to reducedness and irreducibility, respectively. 
For a simple example, consider $f(X)=X^2+1\in\mathbf{Q}[X]$. This is a prime ideal, i.e., $\mathrm{Spec}(\mathbf{Q}[X]/(f(X)))$ is integral. But if we base change to $\mathbf{C}$, we have $f(X)=(X-\sqrt{-1})(X+\sqrt{-1})$, and accordingly $\mathrm{Spec}(\mathbf{C}[X]/(X^2+1))=\mathrm{Spec}(\mathbf{C})\coprod\mathrm{Spec}(\mathbf{C})$ is not connected (though it is reduced).
The only way for reducedness to be lost in an extension of the base field is if that extension is inseparable. So for example take $k=\mathbf{F}_p(T)$, and the polynomial $f(X)=X^p-T\in k[X]$. This is irreducible, so the spectrum of $K=k[X]/(f(X))$ is integral. But if we base change to $K$, the ring we get is $K[X]/(X^p-T)=K[X]/((X-T^{1/p})^p)$, which is not reduced. It is however connected, and this is typical, because, generally, purely inseparable extensions of the base field do not change topological properties. More precisel, if $X$ is a $k$-scheme ($k$ any field), and $K/k$ is a purely inseparable extension, then the projection $X\times_kK\rightarrow X$ is a homeomorphism. 
