I am looking at irreducible algebraic sets.
$V \subseteq K^n$ is an algebraic set $\Leftrightarrow$ it is of the form $V(I)$, where $I=$radical Ideal of $K[x_1, x_2, \dots , x_n]$.
At my lecture notes there is the following:
$V((x^2-y^2))=V_1 \cup V_2$, where $V_1=V(x-y)$ and $V_2=V(x+y)$ algebraic sets.
How do we know that these two are algebraic sets??
Do we have to show that the ideals $\langle x-y \rangle$ and $\langle x+y \rangle$ is radical? How can we do this?
Normally, one does not define algebraic sets associated to radical ideals only, but rather defines them for any ideal or even for any set, like $V(S)= \{p \ \colon p(s) = 0 \ \forall s \in S\}$.
Then, one shows that every algebraic set is in fact already associated to the algebraic set given by some radical ideal. Since $V(S) = V(\langle S \rangle)$ and $V(I)= V (\sqrt{I})$.
Thus, usually, I would say, no you do not have to show that these ideals are radical just to talk about the respective $V$'s. You should check what was the actual definition of $V$ you saw [added: from Proof of the proposition $V(S)=V(\langle S \rangle )$ it appears my guess was correct].
But, also, yes the ideals are radical. This is not hard to see in thise case. They are principal ideals, and the generating element is irreducible so they are radical and even prime ideals.