Questions about field extensions and irreducibility These questions are related to Galois theory, but are general field theory questions. 
First question:
I understand Eisenstein's criterion for proving irreducibility of polynomials over $Q$. However, I can't understand, for example, how to prove irreducibility over $Q(\sqrt{3})$, or any other simple extension. Do I have to attempt to write a potential zero $\alpha$ of the polynomial in the form $a+b\sqrt{3}=\alpha$ and derive a contradiction? Or is there some method by which irreducibility over $Q$ yields information about irreducibility over $Q(\sqrt{3})$ more readily than it appears?
Second question:
Given a polynomial, how do I go about proving that an element is not in its splitting field? Is this again a proof by contradiction by writing the element in question as a linear combination of the spanning vectors of the splitting field over the base field? 
Third question:
This is an incredibly basic question from a previous unit that I still can't seem to wrap my mind around. Suppose I wanted to prove that $\sqrt{5}$ is not contained in $Q(\sqrt{3})$. I assume this would proceed as an irrationality of 2 proof would -- writing: Let $$\sqrt{5} = a + b\sqrt{3}.$$ and show this yields a contradiction. But how? And is there a more general way to do this with any element?
I have a feeling most of these questions are related, since all seem to come down to deriving a contradiction by considering the field extension as a vector space over the base field. If someone could shed some light into this basic concept I would be immeasurably grateful and significantly more prepared to tackle the more complicated Galois theory that is coming my way.
Cheers,
Flabbergasted in Fields
 A: The basic answer to all three questions is that, at this stage in your learning process, you try to write down a general factorisation or expression of an element in terms of basis elements of the field, and derive a contradiction, exactly like you said. There are a couple of points/caveats that are worth mentioning:


*

*Re 1st question: for polynomials of degree greater than 3, reducible is not the same as having a zero over the field. You may have to write down general factorisations into factors of various different degrees. E.g. if your polynomial is of degree 4, you have to show that it has no root and that it doesn't factor into two quadratics over your field.

*There is an Eisenstein Criterion over arbitrary number fields, but you don't yet have the language/background knowledge to state it.

*Sometimes, proving that an element is not in a given field $K$ can be simpler: first you find its minimal polynomial over $\mathbb{Q}$, e.g. by first writing down some polynomial that kills the element, then factorising it and determining which of the irreducible factors kills the element. Now, you know that adjoining this one element to $\mathbb{Q}$ produces an extension of degree equal to the degree of its minimal polynomial, $d$, say. If you get lucky, $d$ will not divide the degree of $K$ over $\mathbb{Q}$, so the element cannot be in $K$ by the tower law. Alternatively, you may find that there are not many subfields of $K$ of degree $d$, so you have to prove that adjoining your element gives you neither of those.

*Later, you will have more ways to prove that two fields (such as $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{5})$) are not equal, using their ramification properties, as one comment notes. For now, you can either solve the obvious equation that you have written down, or use tricks like the following: you basically want to prove that $\mathbb{Q}(\sqrt{3})$ is not the same as $\mathbb{Q}(\sqrt{3},\sqrt{5})$. So you will be done if you can find an element of $\mathbb{Q}(\sqrt{3},\sqrt{5})$ with minimal polynomial of degree 4. Do you have a guess which element to try?
