Isomorphism of algebras $\mathbb{Q}(\cdot , +, 1)$ and $\mathbb{Z}(\cdot , +, 1)$ I have these two algebras and I need to know if they are isomorphic:
$\mathbb{Q}(\cdot , +, 1)$ and $\mathbb{Z}(\cdot , +, 1)$
Are there some general tricks how to deal with this type of tasks?
 A: In this case the two algebras are not isomorphic. The way you prove this is to assume there is an isomorphism $\Phi$ from $\mathbb Z$ to $\mathbb Q$ and then use the fundamental properties of ring isomorphisms:
(1) $\Phi (a+b) = \Phi(a) + \Phi(b)$
(2) $\Phi(ab) = \Phi(a)\Phi(b)$
to show that this map is uniquely determined. Then ask yourself whether this uniquely determined map is in fact an isomorphism.
Some general tips involve looking for "special" elements. If we were looking at group isomorphisms we might prove one group has an element of order $8$ while the other cannot. With infinite algebras we will have to try a little harder. Intuitively each element of $\mathbb Z$ is of the form $\pm(1+1+...+1)$ while the same cannot be said for $\mathbb Q$. Can you use isomorphisms to make this observation rigorous? 
An observation that displays why $\mathbb R$ and $\mathbb Q$ are not isomorphic, is that equations like $x^2 = (1+1)$ have solutions in $\mathbb R$ while not in $\mathbb Q$. In other words $\sqrt2$ is a real number but not a rational. 
You could do the same thing for $\mathbb C$ and $\mathbb Q$ using the equation $x^2 +1 = 0$. Clearly $i^2 + 1 = 0$. Assuming there is an isomorphism $\Phi $ between $\mathbb C$ and $\mathbb Q$ we would then have $\Phi(i^2 + 1) = \Phi(0)$ and therefore $\Phi(i)^2 + 1 = 0$. But there is no rational number $\Phi(i)$ which acts like this, so we cannot have an isomorphism.
Ultimately you want to find a way in which some of one algebra's elements interact, and show none of the other's elements display the same behaviour.
A: Even if we forget the addition and just consider the monoids $(\mathbb{Z},\cdot,1)$, $(\mathbb{Q},\cdot,1)$ there cannot be an isomorphism between them. To see this, notice that in $\mathbb{Z}$ there are only two invertible elements, while there are infinitely many in $\mathbb{Q}$.
You may also forget the multiplication and the unit and show, that $(\mathbb{Z},+)$ and $(\mathbb{Q},+)$ are not isomorphic as groups ($\mathbb{Z}$ is cyclic, $\mathbb{Q}$ is not / $\mathbb{Q}$ is divisible, $\mathbb{Z}$ is not).
This is the common way to show, that some algebraic structures are not isomorphic: just find an algebraic property which is satisfied by exactly one of them and use the fact, that algebraic properties (like the number of invertible elements or being cyclic/divisible/etc.) are preserved by isomorphisms.
