Showing $\mathbf Q(\sqrt2,\mathrm i)=\mathbf Q( \sqrt2+\mathrm i)$. How can we prove that $\mathbf Q(\sqrt2,\mathrm i)=\mathbf Q( \sqrt2+\mathrm i)$?
I have never seen the $\mathbf Q(\sqrt2, \mathrm i)$ notation before, so I am confused as to what it means - is it field generated by $\sqrt{2}$ and $\mathrm i$ over the rationals? How would one write the elements of this field in general terms?
 A: Although a concrete description of the numbers in $\mathbb{Q}(\sqrt{2},i)$ can be given, deriving this representation takes some background work.  I will start from a more abstract definition of these field extensions of $\mathbb{Q}$ and develop the equality posed in the Question.
Intuitively we would like to define $\mathbb{Q}(\sqrt{2},i)$, resp. $\mathbb{Q}(\sqrt{2}+i)$, as the smallest field extension of $\mathbb{Q}$ that contains the indicated numbers.  Note that $\sqrt{2},i$ don't exist in $\mathbb{Q}$, so we need fields that are strictly bigger than $\mathbb{Q}$.
We know one such field, the complex numbers $\mathbb{C}$, which extends $\mathbb{Q}$ and where $\sqrt{2},i$ exist.  We will begin by using that framework, and after the required result:
$$ \mathbb{Q}(\sqrt{2},i) = \mathbb{Q}(\sqrt{2}+i) $$
is obtained, then we will circle back and sketch an argument that, since the same result follows for whatever larger field is used, the conclusion doesn't depend on this choice of "scaffold".
Consider all the fields $\mathbb{F} \subseteq \mathbb{C}$ that are field extensions of $\mathbb{Q}$.  Intersecting all those $\mathbb{F}$ which contain the complex number $\sqrt{2}+i$ gives us the smallest such subfield of $\mathbb{C}$, and this we call $\mathbb{Q}(\sqrt{2}+i)$.  Some details are being omitted, for convenience, but the intersection of a collection of fields that extend $\mathbb{Q}$ will (1) be a field and (2) and extend $\mathbb{Q}$.
Since $\mathbb{Q}(\sqrt{2},i)$ is now a field extension of $\mathbb{Q}$, it is one of the field extensions that contains $\sqrt{2}+i$ (since fields are rings and closed under addition).  Immediately we have:
$$ \mathbb{Q}(\sqrt{2},i) \supseteq \mathbb{Q}(\sqrt{2}+i) $$
because the left hand side is one of the fields $\mathbb{F}$ we intersect to get the right hand side.
To get the reverse inclusion takes a bit of algebra.  Specifically, by taking the reciprocal of $\sqrt{2} + i$ in $\mathbb{C}$:
$$ (\sqrt{2} + i)(\sqrt{2} - i) = 2+1 = 3 $$
$$ \frac{1}{\sqrt{2} + i} = \frac{\sqrt{2} - i}{3} $$
we see that $\sqrt{2} - i = 3/(\sqrt{2} + i)$ also belongs to $\mathbb{Q}(\sqrt{2}+i)$.
Taking the sum and difference of $\sqrt{2} + i$ and $\sqrt{2} - i$ shows that $2\sqrt{2}$ and $2i$ belong to $\mathbb{Q}(\sqrt{2}+i)$.  Multiplying by one-half then proves:
$$ \mathbb{Q}(\sqrt{2},i) \subseteq \mathbb{Q}(\sqrt{2}+i) $$
This completes the proof of equality of these two fields.  It was a choice to carry out the argument in the context of the complex field $\mathbb{C}$, where the meanings of $\sqrt{2}$ and $i$ are familiar.  However the same argument can be carried out in a smaller field extension of $\mathbb{Q}$, for example the algebraic closure of $\mathbb{Q}$ which is a countable field (versus the uncountable cardinality of $\mathbb{C}$).  
The argument could also be carried in a bigger context, but none of the essential details of the proof would need to change.  Therefore the choice of $\mathbb{C}$ can be replaced, and in this sense our conclusion is independent of that choice.

Added:  We can say a little about the "concrete description" of elements of $\mathbb{Q}(\sqrt{2},i)$.  Since $\mathbb{Q}$ is a subfield, $\mathbb{Q}(\sqrt{2},i)$ is a vector space over the rationals.  That is, there exists a basis $\mathscr{B}$ with respect to which every element of $\mathbb{Q}(\sqrt{2},i)$ can be uniquely represented by taking linear combinations using rational coefficients.
It should be fairly well known from high school algebra that $\mathbb{Q}(i)$ is a two-dimensional vector space over $\mathbb{Q}$, and a basis of size 2 for this is $\{1,i\}$, so that everything in $\mathbb{Q}(i)$ can be written with rational real and imaginary parts:  $a + bi$ for certain $a,b \in \mathbb{Q}$.  
It is clear that this descriptive set is closed under addition/subtraction, and (using $i^2 = -1$) under multiplication as well.  The only operation whose closure is less than obvious is division, but recall:
$$ (a + bi)(a - bi) = a^2 + b^2 $$
so that:
$$ \frac{1}{a+bi} = \frac{a-bi}{a^2 + b^2} $$
Since division is the same as multiplying by the reciprocal, this shows the set is closed under divsion also, and because $\mathbb{Q}(i)$ is the smallest field extension of $\mathbb{Q}$ that contains $i$, this set must be all of $\mathbb{Q}(i)$.
A similar representation for $\mathbb{Q}(\sqrt{2})$ can be given by a basis $\{1,\sqrt{2}\}$, i.e.:
$$ \mathbb{Q}(\sqrt{2}) = \{ a + b\sqrt{2} \;\mid \; a,b \in \mathbb{Q} \} $$
Addition/subtraction is obvious (the same for field $\mathbb{Q}(\sqrt{2})$ as for it as the vector space over $\mathbb{Q}$), and multiplication can be carried out using $\sqrt{2}^2 = 2$.  Finally division can be performed using the "conjugate radical" trick:
$$ (a + b\sqrt{2})(a - b\sqrt{2}) = a^2 - 2b^2 $$
where we need the extra fact that $a^2 - 2b^2$ is not zero for rational $a,b$ unless both $a = b = 0$.  This follows from the Fundamental Theorem of Arithmetic, i.e. that prime factorizations are unique (up to commuting factors), which is related to $\sqrt{2}$ being irrational.
The interested Reader is left with an exercise, to show that $\mathbb{Q}(\sqrt{2},i)$ has a basis $\{1,i,\sqrt{2},i\sqrt{2}\}$, and to explain how arithmetic operations can be done with a basis representation of numbers $a + bi + c\sqrt{2} + di\sqrt{2}$ where $a,b,c,d \in \mathbb{Q}$.
