Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to prove that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic. I was thinking of arguing the following:

Suppose there exists an isomorphism $\varphi: \mathbb{Z}[x]\rightarrow\mathbb{Q}[x]$. Because isomorphisms are by definition surjective, there exist $x, y \in\mathbb{Z}[x]$ such that $\varphi(x) = c \in \mathbb{Q}[x]$ and $\varphi(y) = d \in \mathbb{Q}[x]$ for any $c, d\in\mathbb{Q}[x]$. Because $\varphi$ is an isomorphism we must have $\varphi(x+y) = \varphi(x) + \varphi(y)$ for all $x, y \in \mathbb{Z}[x]$. Namely, because polynomial addition is defined componentwise, we must have that the constant term of $\varphi(a + b) = c_{0} + d_{0}$ (where $c_{0}, d_{0}$ are the constant terms of $c$ and $d$ respectively. I would then argue that because $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic as additive groups, no such isomorphism $\varphi$ exists. Is this a valid proof?

I've seen proofs that argue that because $\varphi(1) = 1$ for any homomorphism we have $1 = \varphi(2(1/2)) = 2(\varphi(1/2))$ so $\varphi(1/2)$ must be contained in $\mathbb{Z}[x]^{\times}$. Then because $\mathbb{Z}[x]^{\times} = \mathbb{Z}^{\times} = \{\pm1\}$ we have $2 \times \pm1 \neq1$, a contradiction. Is this any different than arguing that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ each have a different number of units?

share|cite|improve this question
Your proof is not valid. – Martin Brandenburg May 5 '13 at 23:58
Could you explain why? – Danny May 6 '13 at 0:02
@Danny… – user33321 May 6 '13 at 0:53
As discussed in Serkan's link, $R[x]\cong S[x]$ is possible even when $R,S$ are not isomorphic as rings. As additive groups, I'm not sure - but even if it were true when speaking about $R\not\cong S$ as additive groups, the fact of the matter is that that step in your proof is highly non-obvious and in need of justification. – anon May 6 '13 at 1:03
You need to mention that you are referring to isomorphism as an additive group. Since isomorphism means a bijection that preserves structure, you need to explicitly say what structure you are talking about. – mez May 6 '13 at 8:08

12 Answers 12

$\mathbb{Z}[x]^{\times}=\{\pm 1\}$ and $\mathbb{Q}[x]^{\times}=\mathbb{Q}^{\times}$. This is because $R[x]^{\times}=R^{\times}$ for integral domains $R$.

share|cite|improve this answer
This is the best proof because it relies only on a simple lemma that $R[x]^* \cong R^*$ for any $R$. – 6005 Jan 9 '15 at 13:34
@C-S: this is a bit out of date, but your claim as stated isn't true. The set of units in a polynomial ring are the polynomials whose constant term is a unit and whose other coefficients are nilpotent. The claim is true in the case of $\mathbb{Z}$ and $\mathbb{Q}$, since neither have any nonzero nilpotent elements. – Alex Wertheim Apr 16 '15 at 3:15
@AlexWertheim Yeah, thanks. I didn't realize that important restriction on $R$ at the time. – 6005 Apr 16 '15 at 12:18
It is often said that "category theory doesn't help to solve explicit problems". Well, here I thought: "Which functor on $\mathsf{CRing}$ could we apply to simplify the problem directly? What about the group of units $(-)^{\times} : \mathsf{CRing} \to \mathsf{Ab}$?" It worked out pretty well. – Martin Brandenburg Jun 29 '15 at 23:10

Although I've not time to read your proof, you could alternatively use that since $\mathbb{Q}$ is a field, $\mathbb{Q}[x]$ is a principal ideal domain whereas in $\mathbb{Z}[x]$ the ideal $(2,x)$ is an example of an ideal that is not principal.

share|cite|improve this answer
I think this is the best argument. In a similar spirit, one can remark $\mathrm{dim}(\mathbb{Q}[x])=1 < 2 = \mathrm{dim}(\mathbb{Z}[x])$. – Martin Brandenburg May 6 '13 at 8:25
@MartinBrandenburg, this is the most complicated argument, too. – Mariano Suárez-Alvarez May 6 '13 at 21:09
@MarianoSuárez-Alvarez: Dearest Mariano, to remove the stigma of most complicated argument, I posted a second answer that I think you will like. – Jason Polak May 7 '13 at 19:40

$2$ is invertible in $\mathbb{Q}[x]$ but not in $\mathbb{Z}[x]$.

Incidentally, this suggests the following salvage of lhf's now-deleted answer: associated to any subring $R$ of a ring $S$ is its "inverse closure" (I don't know if there's standard notation for this), given by the smallest subring of $S$ containing $R$ and the inverses of every element of $R$ existing in $S$. Given any ring, we can consider the inverse closure of its prime subring, which is its smallest inverse-closed subring. The inverse closure of the prime subring of $\mathbb{Z}[x]$ is $\mathbb{Z}$ while the inverse closure of the prime subring of $\mathbb{Q}[x]$ is $\mathbb{Q}$.

share|cite|improve this answer

The abelian group underlying $\mathbb Q[Z]$ is divisible while that of $\mathbb Z[X]$ is not, so they are not isomorphic even as abelian groups!

share|cite|improve this answer

Suppose $\Bbb{Q}[x]$ and $\Bbb{Z}[x]$ are isomorphic as rings via some map $f$. Then the isomorphism descends into an isomorphism on the quotients $\Bbb{Z}[x]/(x)$ and $\Bbb{Q}[x]/\bigl(f(x)\bigr)$. Now $\bigl(f(x)\bigr)$ is a non-zero prime ideal of $\Bbb{Q}[x]$ and thus is maximal. But now this means that $\Bbb{Z}$ is isomorphic to a field, contradiction.

share|cite|improve this answer
This is incorrect. Why would an isomorphism have to send $x$ to $x$? – KCd May 6 '13 at 3:05
@KCd I have edited my answer. – user38268 May 6 '13 at 3:28
It seems to work now. – Karl Kronenfeld May 6 '13 at 3:45
@Martin: Why do you prefer $\big(f(x)\big)$ over $f((x))$? – Karl Kronenfeld May 6 '13 at 8:49
First I thought that it is a typo and that $f((x))$ doesn't make sense, but of course it makes sense and coincides with $(f(x))$ ... sorry. BenjaLim, feel free to rollback. – Martin Brandenburg May 6 '13 at 9:35

The ring $\mathbb Z[X]$ is finitey generated as a unital ring —by $X$, in fact.

On the other hand, you can easily check that $\mathbb Q[X]$ is not finitely generated as a ring.

share|cite|improve this answer

Lemma: Let $R$ be a ring (commutative with $1$). Then $R$ is a field if and only if $R[X]$ is a PID.

Proof: If $R$ is a field, $R[X]$ is Euclidean and hence a PID. If $R[X]$ is a PID, then since non-zero prime ideals in PIDs are maximal, $\dfrac{R[X]}{(X)} = R$ is a field.

So we need simply observe that $\mathbb{Z}$ is not a field while $\mathbb{Q}$ is.

share|cite|improve this answer

$\mathbb{Z}[x]$ admits quotients of positive characteristic whereas $\mathbb{Q}[x]$ doesn't.

share|cite|improve this answer
Why not a single answer for three proofs? They are all really short ... – Martin Brandenburg May 6 '13 at 8:02
Were we not voting on proofs? I was following @Mariano's lead. – Qiaochu Yuan May 6 '13 at 8:06
What's the point of collecting upvotes?! – Martin Brandenburg May 6 '13 at 9:36
@Martin: who's collecting upvotes? I thought we were ranking proofs. – Qiaochu Yuan May 6 '13 at 17:59

The fundamental theorem of algebraic $K$-theory tells us that $K_*(R[X])\cong K_*(R)$ when $R$ is either $\mathbb Q$ or $\mathbb Z$, because these two rings are regular. The localization theorem for $K$-theory applied to Dedekind domains, and then specialized to $\mathbb Z$, then gives us a long exact sequence that looks like $$\cdots K_{i+1}(\mathbb Q)\to\bigoplus_{\text{$p$ prime}}K_i(\mathbb F_p)\to K_i(\mathbb Z)\to K_i(\mathbb Q)\to\cdots.$$ In particular, using the results of the computation done by Quillen of the higher $K$-theory of finite fields, we get an exact sequence $$0\to K_2(\mathbb Z)\to K_2(\mathbb Q)\to\bigoplus_{\text{$p$ prime}}\mathbb F_p^\times\to\{\pm1\}$$

Since the group $K_{4k-2}(\mathbb Z)$ is finite of order equal to $2c_k$ whenever $k$ is odd and $c_k$ is the numerator of $B_k/4k$, with $B_k$ the $k$-the Bernoulli number, we see that $K_2(\mathbb Z)\cong\mathbb Z/2\mathbb Z$, and this together with the last exact sequence shows that $K_2(\mathbb Z)\not\cong K_2(\mathbb Q)$.

This shows what we wanted.

share|cite|improve this answer
One can always over complicate things! :-) – Mariano Suárez-Alvarez May 7 '13 at 20:18
This is a beautiful proof :) – Jason Polak May 7 '13 at 20:48
Biggest thermonuclear weapon.... – user38268 May 8 '13 at 1:54

Let $\varphi : \mathbb{Z}[x] \to \mathbb{Q}[x]$ be a homomorphism. We claim that $\varphi$ cannot be surjective. To see this, let $\varphi(x) = f$. Then the image of $\varphi$ consists of integer polynomials of $f$. In particular, no element of the image can have a coefficient with a denominator divisible by a prime which doesn't appear in the denominators of the coefficients of $f$. For example, if $f = \frac{x}{2} + \frac{x^2}{3}$, then no element of the image can have a coefficient with a denominator divisible by $5$.

(This argument shows that $\mathbb{Q}[x]$ cannot be generated by one element, so it's closely related to Mariano's answer. It can be straightforwardly generalized to prove the claim in Mariano's answer.)

share|cite|improve this answer

In good-natured response to Mariano's challenge that my proof was the most complicated: We first use that the global dimension of a commutative ring $R$ can be computed by $\mathrm{sup}\{\mathrm{proj.dim}(R/I)\}$, the supremum being taken over all ideals of $R$, and where $\mathrm{proj.dim}(R/I)$ is the projective dimension of $R/I$, i.e. the minimum length projective resolution of $R/I$ as an $R$-module.

The ring $\mathbb{Q}$ is a field and so has global dimension zero, whereas it's easy to see that $\mathbb{Z}$ has global dimension one via the above. We now use one of the first results in dimension theory, which states that $R[x]$ has global dimension $\mathrm{gl.dim}(R) +1$, and so $\mathbb{Z}[x]$ has global dimension two, whereas $\mathbb{Q}[x]$ has global dimension one.

This has the additional property that shows that if $R$ and $S$ have different global dimensions, then $R[x]$ and $S[x]$ cannot be isomorphic.

share|cite|improve this answer

The algebraic closure of the prime ring in $\mathbb Z[X]$ is $\mathbb Z$ while the same thing in $\mathbb Q[X]$ is $\mathbb Q$.

(This is one way to salvage an argument that was given before in a now deleted answer)

share|cite|improve this answer
Er... I think the integral closure of the prime ring in $\mathbb{Q}[x]$ is $\mathbb{Z}$. – Qiaochu Yuan May 7 '13 at 4:23
I meant the algebraic closure, really :-) – Mariano Suárez-Alvarez May 7 '13 at 5:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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