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Given a commutative ring $R$, and $p(x),q(x) \in R[x]$ monic polynomials, under what conditions on $p(x)$ and $q(x)$ are the principal ideals $\langle p(x) \rangle$ and $\langle q(x) \rangle$ isomorphic?

In particular, I wonder how one might classify simple integral extensions $R[x]/I$ of $R$, where $I$ is a principal ideal of $R$. In the case where $R = \mathbb{R}$, for example, if we restrict ourselves to the case of two dimensional extensions over $\mathbb{R}$, there are only three isomorphism classes of rings of this form, $\mathbb{R}[x]/\langle x^2 \rangle, \mathbb{R}[x]/\langle x^2 - 1 \rangle,$ and $\mathbb{R}/\langle x^2 + 1 \rangle$, which although this is a very small example, this makes me wonder whether if some similar behavior to this holds in general, perhaps related to the polynomial discriminant of the polynomials $p(x)$ and $q(x)$ for example.

Are results anything like this known in classifying isomorphic principal ideals of a ring like this, and hence classifying ring extensions of the form $R[x]/\langle p(x) \rangle$ where again $p(x)$ is monic?

(That is, if I'm not butchering my terminology here, the simple integral extensions of $R$.) In particular, I am interested in the case where $R$ is both an Artinian and Noetherian ring, if that is at all relevant.

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Knowing isomorphism classes of ideals of a ring $A$ doesn't give too much information about quotients $A/Ax$. This is the whole point of say the classification of finitely generated modules over PIDs: in a PID every principal ideal is a free $A$-module of rank one (in particular, every two nontrivial ideals of a PID are isomorphic as $A$-modules) but the quotients $A/Ax$ vary. For example, $(2)$ and $(3)$ are isomorphic ideals of $\mathbb Z$ but their quotients do not yield isomorphic rings.

In a mildly general setting, suppose $A$ is any ring and $x\in A$. There is a left $A$-linear map $A\to Ax$ that sends $1\to x$, and this is surjective with kernel the left annihilator of $x$. If $x$ is not a zero divisor, then this is in fact an isomorphism. Thus any left principal ideal not generated by a left zero divisor is a free module of rank one.

In your example, $\mathbb R[X]$ is a PID, and every polynomial in $\mathbb R[X]$ splits as a product of irreducible factors of degree one and two. The quotient ring will contain as many copies of $\mathbb C[X]/(X^q)$ as there are irreducible factors of degree two with multiplicity $q$, and many copies of $\mathbb R[X]/(X^q)$ as there are irreducible factors of degree one with multiplicity $q$.

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I'll answer to the first question.

Since $p(x)$ and $q(x)$ are monic, they're also non-zero divisors, and the ideals $p(x)R[x]$ and $q(x)R[x]$ are both isomorphic to $R[x]$.

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    $\begingroup$ Why the down votes? $\endgroup$ – Bernard Jul 23 '16 at 15:18

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