# 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 $$S \supseteq R?$$

• Nice question! Here's a sort of stupid, trivial example for your more general version: take $R = k[X, Y]$ for $k$ some field, $S = k[X, Y, X^{-1}, Y^{-1}]$, and let $P = \langle X \rangle, Q = \langle Y \rangle$. Clearly, $P \neq Q$, but $PS = QS = S$, since they are generated by units in $S$. (Maybe you want to require that $PS$ and $QS$ remain proper or something. But even then, it seems like taking the same example with $S = k[X, Y, Y/X, X/Y] \subset k(X, Y)$ will work!) Commented Feb 25, 2023 at 6:22
• If you relax the condition that $P, Q$ be primes of $R$ to merely insisting that $P, Q$ be radical ideals of $R$, then if I'm not mistaken, you get an interesting example $R = k[X^{2}, X^{3}], S = k[X], P = X^{2}R, Q = X^{3}R$, where $k$ is again a field. Geometrically, the inclusion $R \to S$ corresponds to the projection of the affine line onto the cuspidal cubic. Commented Feb 25, 2023 at 6:30
• For the first question use that an irreducible (monic) polynomial is uniquely determined by any of its root in the algebraic closure. For the second question what about $R=\Bbb{Z},S=\Bbb{Q},\mathfrak{p}=(2),\mathfrak{q}=(3)$. Commented Feb 25, 2023 at 11:11
• @AlexWertheim Yeah makes sense. Maybe I'll think of a better formulation that makes this non-trivial. Commented Feb 26, 2023 at 5:36
• @reuns I might be acting silly but I don't get your argument for the first one. Yes, the roots determine it, but what if you have repeated roots? For example what if one is $f^4$ and the other $f^2g^2$ in the field extension but irreducible in the ground field? Commented Feb 26, 2023 at 5:39

$$\DeclareMathOperator{rad}{rad}$$ 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 $$S$$.
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.