How to show $\Bbb Q$ and $\Bbb R$ are not isomorphic, what about $\Bbb R$ and $\Bbb C$ - what are some field isomorphism invariants? This is probably a dumb question(and I can't find a previous iteration of this question), but...
How do I rigorously show that $\Bbb Q$ and $\Bbb R$ are not isomorphic? I mean there is a notion of $\Bbb Q$ being smaller than $\Bbb R$, and perhaps it has something to do with $\sqrt{2}$ being in $\Bbb R$ and not in $\Bbb Q$ which makes me think of roots of polynomials, but I don't know how to rigorously show it at all. 
I suppose my question is two fold, how does one show that $\Bbb Q$ and $\Bbb R$ aren't isomorphic, but more generally, what are 'field' isomorphism invariants? What properties can one use to determine immediately that two fields aren't isomorphic?
Take $\Bbb R$ and $\Bbb C$ as well.
 A: Well, if you have a field isomorphism $\phi: F\to K$, then if we have a polynomial equation
$$
x^n + a_{n-1} x^{n-1}+ \ldots + a_0 = 0,
$$
with coefficients in $F$, we get an equation
$$
x^n + \phi(a_{n-1})x^{n-1} + \ldots \phi(a_0) = 0
$$
with coefficients in $K$. Then if $\alpha$ is a solution to our equation with coefficients in $K$, $\phi^{-1}(\alpha)$ solves the equation in $F$. 
Then we can use this to see that $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are all non-isomorphic, since if we have a field isomorphism $\phi$ from $\mathbb{Q}$ to $\mathbb{R}$ or $\mathbb{R}$ to $\mathbb{C}$ then surely $\phi(2) = 2$ and $\phi(-1)= -1$.
But then the equations $x^2 -2=0$ and$x^2 + 1=0$ would have solutions in all the fields, and we can see that this isn't the case. 
So, looking at equations which have solutions in some fields but not others can be used to show that fields are not isomorphic.
A: I think size is the easiest requirement for two things to be isomorphic. Just as no group with 5 elements can be isomorphic to a group with 2 elements, an uncountable field cannot be isomorphic to a countable field.
A: Like it is said in the comment, for the rational field and real field, clearly it is the cardinality that easily allows you to give a contradiction. 
However, if you want a more algebraic way to show that $\mathbb{Q}$ is not isomorphic to $\mathbb{R}$ you can consider the following, there exists a polynomial of degree $3$ over $\mathbb{Q}$ which has no root ($X^3-2$ for instance) whereas any polynomial of order $3$ over $\mathbb{R}$ has a root (it follows from the intermediate value theorem).
Now for $\mathbb{R}$ and $\mathbb{C}$ any non-constant polynomial over $\mathbb{C}$ has a root whereas the polynomial $X^2+1$ in $\mathbb{R}$ has no root.
A little more funny, if you admit the axiom of choice then $\mathbb{C}_p$ is isomorphic to $\mathbb{C}$ but there are no concrete isomorphism that one can construct to see this. It means, IMHO, that purely algebraic isomorphisms become a little messy when we have high cardinality (here the topologies on both fields are radically different). 
A: Well, size is indeed a point: an isomorphism $f$ between field $K$ and $L$ is in particular a bijection, so by definition $|K| = |L|$. The fun for finite fields is that the reverse holds: fields of the same finite size are isomorphic!
This is a particularly easy to way to distinguish the rationals and the reals.
But the $\sqrt{2}$ idea also works: the field $\mathbb{R}$ has the field property 
$$\exists x \in K: x \cdot x = 1 + 1\text{.}$$ If $f: K \rightarrow L$ is an isomorphism then if $a \in K$ is such an element $a$ that obeys it, then $f(a) \cdot f(a) = f(a \cdot a) = f(1+1) = 1+1$, so $f(a)$ fulfills the property in $L$ as well.
The property above is a first order property (expressible in quantifiers and the field operations and constants) that holds in a field. If it holds in a field, it holds in isomorphic fields as well (similarly to the $\sqrt{2}$-property above).
E.g. the complex numbers can be distinguished from the reals because in the complex numbers we have $$\exists x \in K: x \cdot x + 1 = 0\text{.}$$ (which corresponds to the idea that we have $i$), and one can show that in the reals no such $x$ can exist. Also higher order properties (like being algebraically closed) can be used.  
