Why the group $(\mathbb{Q}[x],+)$ is not isomorphic to $(\mathbb{Z}[x],+)$ or even $(\mathbb{Q},+)$?

I have been asked the question "Why is the group $(\mathbb{Q}[x],+)$ not isomorphic to $(\mathbb{Z}[x],+)$ or even $(\mathbb{Q},+)$?" . I will be thankful for any help. :)

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It would radically facilitate answering your question if you could indicate what makes you think they might be isomorphic. – joriki Apr 24 '11 at 17:23
After reading the answers, I think I may have misunderstood your question. Did you mean to ask "How can we prove that $(\mathbb Q[x],+)$ is not isomorphic to $(\mathbb Z[x],+)$ or even $(\mathbb Q,+)$"? This is the question that the answers are answering. When you ask "Why are they not isomorphic?", it sounds as if there's some reason to believe that they might be. – joriki Apr 24 '11 at 19:35
@joriki: Thanks for your comments. Yes, Matt was right,this is a homework one. But I added it just to verify that in another way, not to get the common strong proofs posed below. I thought about the first part Q[x] and Z[x] as an infinite abelian groups and the second part, Q[x] and Q. Maybe, I asked the question in an improper way. I am probing another pretty way in which, this question could be justified. For example, with the help of group extentions which is prime tools for investigating the infinite abelian groups. Thanks for consideration. :) – Babak S. Apr 25 '11 at 11:53

If ${\mathbb Q}$ were isomorphic to ${\mathbb Q}[x]$, then there would be an isomorphism $\phi$ taking ${\mathbb Q}$ to ${\mathbb Q}[x]$. The behavior of $\phi$ on ${\mathbb Q}$ is dictated by what $\phi(1)$ is: if $m$ is a natural number then $$\phi(m) = \phi(1 + ... +1) = \phi(1) + .... + \phi(1) = m\phi(1)$$ Then since $0 = \phi(0) = \phi(-m + m) = \phi(-m) + \phi(m)$, you get $\phi(-m) = -m\phi(1)$. Similarly, if ${a \over b}$ is any rational number you have $$a\phi(1) = \phi(a) = \phi({a \over b} + ... + {a \over b}) = b\phi({a \over b})$$ So $\phi({a \over b}) = {a \over b}\phi(1)$. This means $\phi({a \over b})$ has the same degree as $\phi(1)$, regardless of what $\phi(1)$ is... so it's not going to be surjective.

As for why ${\mathbb Q}[x]$ is not isomorphic to ${\mathbb Z}[x]$, notice ${\mathbb Q} \subset {\mathbb Q}[x]$ when viewed as constant polynomials. So if there were an isomorphism $\psi$ taking ${\mathbb Q}[x]$ to ${\mathbb Z}[x]$, $\psi$ would take ${\mathbb Q}$ isomorphically to some subset of ${\mathbb Z}[x]$. But then you can argue like above to say $\psi({a \over b}) = {a \over b} \psi(1)$. (Technically you prove $b \psi({a \over b}) = a\psi(1)$). Thus if you take $b$ large enough, no matter what $\psi(1)$ is, the polynomial $\psi({a \over b})$ cannot have integer coefficients.

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$\mathbb{Q}[X] \not\cong \mathbb{Z}[X]$ because $\mathbb{Q}[X]$ has infinitely many units, where as $\mathbb{Z}[X]$ has only finite number of units. – user9413 Jun 1 '11 at 15:41
@user9413: Units are part of the ring structure; the question is about the additive group structure, and a homomorphism of the additive groups don't need to respect the multiplicative structure. – Hurkyl Feb 15 '13 at 8:42

I'm presuming this is homework from the wording, but I didn't want to add the tag without verification; still, these will be hints rather than spoilers.

For the first half: what properties do you know that $\mathbb{Q}$ satisfies as a group that $\mathbb{Z}$ doesn't? Can you find one that $\mathbb{Q}[x]$ satisfies, and show that $\mathbb{Z}[x]$ doesn't? (Slightly bigger hint: what does it mean for a number to be 'even', and why doesn't that concept make sense in $\mathbb{Q}$?)

For the second half: what binary operations can you perform on (nonzero) members of $\mathbb{Q}$ that are guaranteed to give you members of $\mathbb{Q}$? Are there any that 'don't work right' when performed on two members of $\mathbb{Q}[x]$?

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I'm not sure I understand the second hint. – Qiaochu Yuan Apr 24 '11 at 17:38
@Qiaochu: He's hinting at division of polynomials. – Brandon Carter Apr 24 '11 at 17:48
@Brandon Carter: True, but then a proof of the non-isomorphism would have to show that no "division" (or more familiarly multiplication) can be defined on $\mathbb{Q}[x]$ that interacts with addition in the same way as normal multiplication/division does in $\mathbb{Q}$. This division/multiplication need not be the "natural" one. That would be a viable approach, but surely not a simplest one. – André Nicolas Apr 24 '11 at 17:56
Depending by his/her background, might be easier to look to them as $\Z$-modules and check bases and linearly independent sets in each of them.... – N. S. Apr 24 '11 at 18:03
The division is pretty straightforward, actually; addition doesn't even come into it. All one needs to do is exhibit two non-zero polynomials $\mathfrak{a}, \mathfrak{b}$ such that $\mathfrak{a}/\mathfrak{b}$ isn't a polynomial, and there are choices for $\mathfrak{a}$ and $\mathfrak{b}$ that make such a proof pretty straightforward (assuming only that the questioner knows how degrees of polynomials behave). – Steven Stadnicki Apr 25 '11 at 0:34

There are many possible approaches to proving that two algebraic structures $\mathbb{A}$ and $\mathbb{B}$ are not isomorphic. One way is to show that $\mathbb{A}$ and $\mathbb{B}$ differ in some structural property that must be preserved by any isomorphism. Most dramatically, $\mathbb{A}$ and $\mathbb{B}$ might be of different sizes. Or else (in groups) $\mathbb{A}$ might have an element of order $2$, while $\mathbb{B}$ does not. Such structural arguments are often quite simple. However, finding the right structural property may require considerable insight or experience.

Another approach, logically not very different from the first, is to assume that the mapping $\phi$ from $\mathbb{A}$ to $\mathbb{B}$ is an isomorphism, and show that something must go bad. We will do a detailed calculation using one of your questions as an example.

Suppose that the mapping $\phi$ from the additive group $\mathbb{Q}$ to the additive group $\mathbb{Q}[x]$ is a group isomorphism. Suppose that this $\phi$ takes the rational $r$ to the polynomial $1$, and takes the rational $s$ to the polynomial $x$.

Let $r=a/b$, and $s=c/d$, where $a$, $b$, $c$, $d$ are integers and $b$, $d$ are positive. Clearly neither $a$ nor $c$ is $0$.

Since $a=r+\cdots +r$ ($b$ times), $\phi$ takes $a$ to the polynomial $b$. And if $a$ is positive, then since $a=1+\cdots+1$ ($a$ times), $\phi$ takes $1$ to the polynomial $b/a$. An easy argument shows that this is also true if $a$ is negative.

An almost identical argument shows that $\phi$ takes $1$ to the polynomial $(d/c)x$. This is impossible: $\phi$ cannot take $1$ to $b/a$, and also take $1$ to $(d/c)x$, since obviously $b/a$ and $(c/d)x$ are distinct polynomials.

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@user6312: consider $p(x)=x^{2}-\frac{1}{4}$. Thus $p(x)$ has a root in $\mathbb{Q}[x]$ where as it doesn't have a root in $\mathbb{Z}[x]$. Does that suffice to prove that they are not isomorphic. – user9413 May 13 '11 at 14:57
@Chandru1: You probably mean $p(x)$ has a root in $\mathbb{Q}$, but not in $\mathbb{Z}$, although your version also makes sense. No, it would not suffice, since in general there is nothing built into the definition of isomorphism that forces the isomorphism to take constant polynomials to constant polynomials. – André Nicolas May 13 '11 at 15:19