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. :)
 A: 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 $\mathbb{Q}[x]$?
A: 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. 
A: 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. 
