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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 $S \supseteq R?$

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  • $\begingroup$ 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!) $\endgroup$ Commented Feb 25, 2023 at 6:22
  • $\begingroup$ 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. $\endgroup$ Commented Feb 25, 2023 at 6:30
  • $\begingroup$ 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 $R=\Bbb{Z},S=\Bbb{Q},\mathfrak{p}=(2),\mathfrak{q}=(3)$. $\endgroup$
    – reuns
    Commented Feb 25, 2023 at 11:11
  • $\begingroup$ @AlexWertheim Yeah makes sense. Maybe I'll think of a better formulation that makes this non-trivial. $\endgroup$ Commented Feb 26, 2023 at 5:36
  • $\begingroup$ @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? $\endgroup$ Commented Feb 26, 2023 at 5:39

1 Answer 1

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$\DeclareMathOperator{rad}{rad}$ 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.

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