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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?

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  • $\begingroup$ Your proof is not valid. $\endgroup$ Commented May 5, 2013 at 23:58
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    $\begingroup$ Could you explain why? $\endgroup$
    – Danny
    Commented May 6, 2013 at 0:02
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    $\begingroup$ @Danny math.stackexchange.com/questions/13504/… $\endgroup$
    – user33321
    Commented May 6, 2013 at 0:53
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    $\begingroup$ 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. $\endgroup$
    – anon
    Commented May 6, 2013 at 1:03
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    $\begingroup$ 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. $\endgroup$
    – mez
    Commented May 6, 2013 at 8:08

13 Answers 13

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

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  • $\begingroup$ This is the best proof because it relies only on a simple lemma that $R[x]^* \cong R^*$ for any $R$. $\endgroup$ Commented Jan 9, 2015 at 13:34
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    $\begingroup$ @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. $\endgroup$ Commented Apr 16, 2015 at 3:15
  • $\begingroup$ @AlexWertheim Yeah, thanks. I didn't realize that important restriction on $R$ at the time. $\endgroup$ Commented Apr 16, 2015 at 12:18
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    $\begingroup$ 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. $\endgroup$ Commented Jun 29, 2015 at 23:10
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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.

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    $\begingroup$ I think this is the best argument. In a similar spirit, one can remark $\mathrm{dim}(\mathbb{Q}[x])=1 < 2 = \mathrm{dim}(\mathbb{Z}[x])$. $\endgroup$ Commented May 6, 2013 at 8:25
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    $\begingroup$ @MartinBrandenburg, this is the most complicated argument, too. $\endgroup$ Commented May 6, 2013 at 21:09
  • $\begingroup$ @MarianoSuárez-Alvarez: Dearest Mariano, to remove the stigma of most complicated argument, I posted a second answer that I think you will like. $\endgroup$
    – user2055
    Commented May 7, 2013 at 19:40
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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!

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$2$ 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 $\mathbb{Q}$.

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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.

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    $\begingroup$ This is incorrect. Why would an isomorphism have to send $x$ to $x$? $\endgroup$
    – KCd
    Commented May 6, 2013 at 3:05
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    $\begingroup$ @KCd I have edited my answer. $\endgroup$
    – user38268
    Commented May 6, 2013 at 3:28
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    $\begingroup$ It seems to work now. $\endgroup$ Commented May 6, 2013 at 3:45
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    $\begingroup$ @Martin: Why do you prefer $\big(f(x)\big)$ over $f((x))$? $\endgroup$ Commented May 6, 2013 at 8:49
  • $\begingroup$ 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. $\endgroup$ Commented May 6, 2013 at 9:35
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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.

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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.

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

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    $\begingroup$ Why not a single answer for three proofs? They are all really short ... $\endgroup$ Commented May 6, 2013 at 8:02
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    $\begingroup$ Were we not voting on proofs? I was following @Mariano's lead. $\endgroup$ Commented May 6, 2013 at 8:06
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    $\begingroup$ What's the point of collecting upvotes?! $\endgroup$ Commented May 6, 2013 at 9:36
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    $\begingroup$ @Martin: who's collecting upvotes? I thought we were ranking proofs. $\endgroup$ Commented May 6, 2013 at 17:59
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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.

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    $\begingroup$ One can always over complicate things! :-) $\endgroup$ Commented May 7, 2013 at 20:18
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    $\begingroup$ This is a beautiful proof :) $\endgroup$
    – user2055
    Commented May 7, 2013 at 20:48
  • $\begingroup$ Biggest thermonuclear weapon.... $\endgroup$
    – user38268
    Commented May 8, 2013 at 1:54
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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.)

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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.

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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)

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  • $\begingroup$ Er... I think the integral closure of the prime ring in $\mathbb{Q}[x]$ is $\mathbb{Z}$. $\endgroup$ Commented May 7, 2013 at 4:23
  • $\begingroup$ I meant the algebraic closure, really :-) $\endgroup$ Commented May 7, 2013 at 5:20
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First method:

Let $\Bbb{Z}[x]$ and $\Bbb{Q}[x]$ be isomorphic.$\\[2ex]$
Let $\phi:\Bbb{Q}[x]\rightarrow\Bbb{Z}[x]$ where $x\mapsto \phi(x)$ be an isomorphism.

We note that $\dfrac{x}{2^n}\in\Bbb{Q}[x] \;\;\forall n\in\Bbb{N}$

Now \begin{align} \phi(x) &= \phi\Big(2^n\cdot\dfrac{x}{2^n}\Big)\\[2ex] &= 2^n\cdot\phi\Big(\dfrac{x}{2^n}\Big)\;\text{[Since $ \phi$ is a homomorphism]}\\[2ex] \end{align}
As $\phi$ is injective, $\phi\Big(\dfrac{x}{2^n}\Big)\ne 0$. Since $\phi\Big(\dfrac{x}{2^n}\Big)$ is a non-zero polynomial with integer coefficients, the absolute values of the non-zero coefficients of $2^n\cdot\phi\Big(\dfrac{x}{2^n}\Big)$ is at least $2^n$.
Since this is true for any $n\in\Bbb{N}$, the coefficients of the polynomial $\phi(x)=2^n\cdot\phi\Big(\dfrac{x}{2^n}\Big)$ is arbitrarily large, which is impossible.

Thus $\not\exists$ any isomorphism between $\Bbb{Z}[x]$ and $\Bbb{Q}[x]$.

Second method:

Let $\phi:\Bbb{Q}[x]\rightarrow \Bbb{Z}[x]$ be an isomorphism. $\\[2ex]$
Since $\phi$ is a ring homomorphism, we have $\phi(1)=1$

Now $1=\phi(1)=\phi\Big(2\cdot\dfrac 12\Big)=2\cdot \phi\Big(\dfrac 12\Big)$, since $\phi$ is a homomorphism.
Since $\phi\Big(\dfrac 12\Big)\in\Bbb{Z}[x]$, we write
$\phi\Big(\dfrac 12\Big)= a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1+a_0$ where $a_i\in\Bbb{Z}, i=0(1)n$
Since $2\phi\Big(\dfrac 12\Big)=1$, it follows that
$2a_n=0,2a_{n-1}=0,\cdots,2a_1=0,2a_0=1\implies a_0=\dfrac 12\not\in\Bbb{Z}$, a contradiction.
So $\Bbb{Z}[x]$ and $\Bbb{Q}[x]$ are not isomorphic.

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