First, I ask my question and then I add some explanations:

Suppose that $A$ and $B$ are two commutative rings such that $A[X] \cong B[X]$ as rings. Denote by $D_A$ the set of all positive integers $n$ such that there exists an irreducible polynomial of degree $n$ over $A$ — the same for $D_B$. Is it true that $D_A = D_B$?

Some time ago, I wanted to find many proofs (like here) that $\Bbb Z[X]$ and $\Bbb R[X]$ are not isomorphic (obviously they are not because they don't even have the same cardinality, I know). I thought to the following argument:

"The irreducible polynomials in $\Bbb R[X]$ have degree $≤2$, while irreducible polynomials in $\Bbb Z[X]$ can have arbitrary large degree (for instance $X^n+2X^{n-1}+\cdots+2X+2$, by Eisenstein's criterion)".

But I wasn't sure of the correctness of this argument. The isomorphism $A[X] \cong B[X]$ is not required to preserve the degree. If it is preserved, then my claim should be true. I think that examples like this could prevent the isomorphism from preserving the degree.

A possibly relevant question is What are the possible sets of degrees of irreducible polynomials over a field?, on MO. In particular, this can be interesting when $D_A$ and $D_B$ are infinite.

Thank you for your comments!

  • 3
    $\begingroup$ An interesting question. At least initially I'm a bit concerned about the possibility of zero divisors. Is the usual definition of irreducible still ok when there are zero divisors around? Over $\Bbb{Z}_4$ (or any ring with nilpotent elements) we have surprise units: $(1+2x)^2=1$, and over $\Bbb{Z}_6$ we have beauties like $(2x+1)(3x+1)=1-x$. OTOH if we assume that $A$ and $B$ are integral domains, then we can use their respective fields of fractions... (though Gauss' lemma may fail). $\endgroup$ Jul 25, 2016 at 20:50
  • 4
    $\begingroup$ An isomorphism $\mathbb{R}[X]\to\mathbb{Z}[X]$ would mean $\mathbb{Z}[X]$ is a PID, which it is not. It's not difficult to show that, if $A$ is a field and $\varphi\colon A[X]\to B[X]$ is an isomorphism, then also $B$ is a field and $\varphi$ induces an isomorphism $A\to B$ (via restriction); moreover $\varphi(X)$ must have degree $1$. Thus $\varphi$ preserves degrees. $\endgroup$
    – egreg
    Jul 25, 2016 at 21:18
  • $\begingroup$ Possibly related: math.stackexchange.com/questions/1675650/… $\endgroup$
    – Watson
    Aug 17, 2016 at 20:25
  • $\begingroup$ If we assume that $A$ is a domain, then the units of $A[X]$ are mapped to units of $B[X]$, i.e. $A^{\times} \cong B^{\times}$. Therefore, the only way for the degree not to be preserved is that a non-unit in $A$ is mapped to a polynomial of degree $ \geq 1 $ in $B[X]$. $\endgroup$
    – Watson
    Feb 3, 2017 at 21:55
  • $\begingroup$ (Of course, $D_A = D_B$ doesn't imply that $A[X] \cong B[X]$, see $A = \Bbb F_2, B = \Bbb F_3$.) $\endgroup$
    – Watson
    Jun 25, 2018 at 7:25

1 Answer 1


Edit2: I'm afraid my argument below breaks down. $B[X]$ may have more maximal ideals than the "uppers" of the $\mathfrak n\in max(B)$, so that the primes of codimension 1 are not necessarily the $\mathfrak n[X]$ with $\mathfrak n\in max(B)$. Here's an example: let $B=\mathbb{Z}_{(2)}=\{m/n|m,n\in\mathbb{Z}\,\&n \not\equiv 0(\textrm{mod}\ 2)\}$, a local domain of dimension 1. We have a surjection $g:B[X]\twoheadrightarrow\mathbb{Q}$ given by $X\mapsto 1/2$. So the kernel of $g$ is a maximal ideal of $B[X]$, while this ideal is the "upper" $<\underline 0,X-1/2>$ to the minimal prime $\underline 0$ of $B$. My apologies... End-of-edit2.

Just a few comments. As is well-known, prime ideals of $A[X]$ come in two flavors, "downers" and "uppers". The former are the $\mathfrak p[X]=\{t\in A[X]|$ all coefficients of $t$ are in $\mathfrak p\}$, and latter are the $<\mathfrak p,h> = \{t\in A[X]|\, t$ mod $\mathfrak p[X]$ is divisible by $h$ in $Q(A/\mathfrak p)[X]\}$ (where $h\in Q(A/\mathfrak p)[X]$ is any monic, irreducible polynomial), with $\mathfrak p\in spec(A)$. One has $\mathfrak p[X] \subsetneq <\mathfrak p,h>$, and both lie over the prime ideal $\mathfrak p$ of $A$.

Now let $f:A[X]\to B[X]$ be a ring isomorphism.

If $\mathfrak m\in max(A)$, the "lower" $\mathfrak m[X]$ is a prime of codimension one in $A[X]$ (only the uppers to $\mathfrak m$ are larger, and there are no inclusion relations between them), so $f(\mathfrak m[X])$ must be of codimension one in $B[X]$, hence of the form $\mathfrak n[X]$ with $\mathfrak n\in max(B)$. This clearly gives a bijection between the maximal ideals of $A$ and those of $B$. And $f$ will then induce an isomorphism $A/\mathfrak m[X]\to B/\mathfrak n[X]$. As $A/\mathfrak m$ and $B/\mathfrak n$ are fields, we find that for each maximal ideal $\mathfrak n$ of $B$ there must be a $h\in \mathfrak n[X]$ and $b,c\in B$ with $b\notin \mathfrak n$ such that $f(X)=bX+c+h$. (In particular, the coefficient of $X$ in $f(X)$ is not in any maximal ideal of $B$, hence it is a unit in $B$.)

But I'm not sure under what circumstances this implies that $f(X)$ must be a linear polynomial.

Edit1: if $rad(B)$ denotes the Jacobson radical of $B$, that is $\bigcap max(B)$, we have a natural injection $(B/rad(B))[X]\rightarrowtail\prod_{\mathfrak n\in max(B)}((B/\mathfrak n)[X])$. Let $c$ be the coefficient of $X^{n}$ in $f(X)$ for some $n\geq 2$; since the image of $f(X)$ in every $B/\mathfrak n[X]$ is linear, it follows that $c\in rad(B)$. In other words, $f(X)$ is the sum of a linear polynomial and a polynomial with coefficients in $rad(B)$. In particular, $f(X)$ is linear when $rad(B)=0$, or, equivalently, $rad(A)=0$.

And if $f(X)=bX+c$ is linear, clearly $b\in B^{*}$, $f(A)=B$, and $f$ preserves degrees - in particular those of the irreducible polynomials over $A$. End-of-edit1.

The minimal prime ideals of $A$ and $B$ must also correspond to each other, as $\mathfrak p[X]\in min(A[X])$ for each $\mathfrak p\in min(A)$, and all minimal primes of $A[X]$ (and $B[X]$) are of this form. Perhaps this observation could be of some use when someone settles the case where $A$ and $B$ are domains.

When they are domains, we cannot conclude $f$ extends to $Q(A)[X]\to Q(B)[X]$ - unless we know that $f(A)\subseteq B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.