Radicals ideals equal in field extension If $f,g \in F[x]$ ($F$ is a field) are two polynomials, then we can obviously have $\text{rad} (f)=\text{rad} (g),$ for instance, $f=x$ and $g=x^2.$ However if the polynomials are irreducible, then this isn't possible.
However, if we consider some field extension $K/F,$ is $\text{rad} (f)=\text{rad} (g)$ possible in $K[x]?$ I am taking $f,g$ to be irreducible in $F[x].$
If $f,g$ are irreducible in a characteristic $0$ field, then $f,g$ won't have repeated factors in $K,$ and so $\text{rad} (f)=\text{rad} (g)$ won't be possible. So I believe my question is more about other fields, where we can have repeated factors, for instance, $x^2+t \in \mathbb{F}_2(t)[x]$ factors as $(x+\sqrt{t})^2$ in $\mathbb{F}_2(\sqrt{t})[x]$   which has repeated factors.
Perhaps a more general version is to ask if for two prime ideals $\mathfrak{p} \ne \mathfrak{q}$ in a ring $R,$ can we have $\text{rad} \; \mathfrak{p} = \text{rad} \; \mathfrak{q}$ in some ring extension $S \supseteq R?$
 A: $\DeclareMathOperator{rad}{rad}$
Suppose $R,S$ are commutative rings and $R$ is a subring of $S$.
To address the original question, suppose $R,S$ are UFDs and $f,g$ in $R$ have distinct radicals. We will show $f,g$ have distinct radicals in $S$ when two assumptions hold: (i) relatively prime elements of $R$ remain relatively prime in $S$ and (ii) non-units of $R$ remain non-units in $S$.  These assumptions both hold when $R=F[x]$, $S=K[x]$ as in the original post.  In fact, (i) holds anytime $R$ is a PID.  (To see this, suppose $a,b\in R$ are relatively prime; there are $\alpha,\beta\in R$ with $\alpha a+\beta b=1$.  This equation is valid in $S$, so $a,b$ are relatively prime there.)  Assumption (ii) holds whenever $S$ is an integral extension or a faithfully flat extension of $R$.
To verify this claim, note that since $f,g$ have different radicals, there is an irreducible $p$ that is a factor of one but not the other, say $p\mid f$, $p\nmid g$.  Thus $p,g$ are relatively prime in $R$ and so in $S$ by assumption (i).  Moreover, $p$ is not a unit in $S$ by assumption (ii).  Thus $p$ has some irreducible factor in $S$ that is not a factor of $g$ (but of course is a factor of $f$).  It follows that $f,g$ have distinct radicals in $S$.
The comments after the original post show that problems arise when condition (ii) fails.
Here is a positive result for the general question asked at the end of the post: if $S$ is an extension of $R$ for which lying over holds (for example, $S$ is an integral extension of $R$ or $S$ is a faithfully flat extension of $R$) and $P,Q$ are distinct prime ideals of $R$, then $\rad PS\ne\rad QS$.  Suppose to the contrary that $\rad PS=\rad QS$.  Lying over means there is a prime ideal $P'$ of $S$ with $P'\cap R=P$.  Clearly $P'$ contains $PS$, so it contains $\rad PS$.  Thus $Q\subseteq QS\cap R\subseteq (\rad QS)\cap R\subseteq P'\cap R=P$.  Reversing the roles of $P$ and $Q$ shows $P=Q$, contrary to our assumption.
