Proving that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.

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?

• Your proof is not valid. – Martin Brandenburg May 5 '13 at 23:58
• Could you explain why? – Danny May 6 '13 at 0:02
• – 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

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

• 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• @C-S: this is a bit out of date, but your claim as stated isn't true.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.

• 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-Álvarez 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. – user2055 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 $\mathbb{Q}$.

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!

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.

• 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

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.

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.

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

• 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.

• One can always over complicate things! :-) – Mariano Suárez-Álvarez May 7 '13 at 20:18
• This is a beautiful proof :) – user2055 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.)

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.

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)

• 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-Álvarez May 7 '13 at 5:20