What does mean of the following field extension? I am working on a problem set and the first question asks:

Give a basis for each of the following vector spaces over the
  indicated fields.
  
  
*
  
*$\mathbb{R}(\sqrt2)$ over $\mathbb{R}$.
  
*$\mathbb{Q}(2^{1/4})$ over $\mathbb{Q}$.
  

Frankly, I have never seen a set denoted as $\mathbb{R(\sqrt2)}$ or $\mathbb{Q(2^{1/4})}$ and therefore I am unable to answer this problem. I checked online and Google gives hits on pdfs that have something to do with "Galois Theory," which I haven't studied either. 
 A: Let $F$ be a field, and let $\Omega$ be a large field containing $F$ as a subfield.  For $S \subseteq \Omega$, define $F(S)$ to be the intersection of all subfields of $\Omega$ containing $F$ and $S$.  And, define $F[S]$ to be the intersection of all subrings of $\Omega$ containing $F$ and $S$.  
Remember that a subfield of $\Omega$ is a nonempty subset which is closed under addition, subtraction, multiplication, and division. A subring is the same, exception division is not required.  
Then $F(S)$ is a field, and $F[S]$ is a ring, because the intersection of a collection of fields (rings) is a field (ring).  There are more concrete descriptions of $F(S)$ and $F[S]$.  The field $F(S)$ is equal to the set of quotients
$$\frac{f(x_1, ... , x_n)}{g(x_1, ... , x_n)}$$
where $n$ is any positive integer, $f$ and $g$ are polynomials in $n$ variables with coefficients in $F$, $x_1, ... , x_n$ are in $S$, and $g(x_1, ... , x_n) \neq 0$.  To see this, note that every subfield of $\Omega$ containing $F$ and $S$ must contain all such quotients.  On the other hand, these quotients themselves form a subfield of $\Omega$ which contains $F$ and $S$, so this field contains $F(S)$, by definition.
Similarly, you can argue that $F[S]$ is equal to the set of elements of the form $f(x_1, ... , x_n)$, where $n\geq 1$, $f$ is a polynomial in $n$ variables with coefficients in $F$, and $x_1, ... , x_n \in S$.
If $S = \{s_1, ... , s_n\}$ is a finite set, then one writes $F(s_1, ... , s_n)$ instead of $F(\{s_1, ... , s_n\})$, and similarly $F[s_1, ... , s_n]$.
Theorem: Assume $s \in \Omega$ is algebraic over $F$, that is, $s$ is the root of a nonzero polynomial $f(X)$ with coefficients in $F$.  Then $F[s] = F(s)$.
Proof: Choose $f$ so that its degree is minimal.  Then $f$ is irreducible in the ring $F[X]$: if $f$ could be written as $f_1f_2$ for some polynomials $f_i$ of strictly lower degree than $f$, then we would have $f_1(s) f_2(s) = f(s) = 0$, hence $f_1(s)$ or $f_2(s) =0$, contradicting the assumption that $f$ is a polynomial of minimal degree such that $f(s) = 0$.
Let $h \in F[X]$, with $h(s) \neq 0$.  Then $f$ cannot divide $h$ in $F[X]$.  Combined with the fact that $f$ is irreducible, we can conclude that the greatest common divisor of $f$ and $h$ is $1$.  By the Euclidean algorithm (or by some general properties of principal ideal domains), there exist polynomials $q_1, q_2 \in F[X]$ such that $q_1h + q_2f = 1$.  Then $1 = q_1(s)h(s) + q_2(s)f(s) = q_1(s)h(s)$, or $q_1(s) = \frac{1}{h(s)}$.
This shows that $F(s)$ is a subset of $F[s]$.  On the other hand, $F[s]$ is obviously a subset of $F(s)$.  Hence $F(s) = F[s]$.  $\blacksquare$
Since $F(s)$ is a field containing $F$, it can naturally be thought of as a vector space over $F$: vector addition in $F(s)$ is the usual addition, and scalar multiplication by $F$ comes from the usual multiplication of elements of $F(s)$.  The following theorem gives you a way to solve your problem.
Theorem: Assume $s$ is algebraic over $F$.  Let $f \in F[X]$ be a polynomial of minimal degree such that $f(s) = 0$.  Let $n$ be the degree of $f$.  Then $F(s)$ is an $n$-dimensional vector space over $F$, and a basis is
$$1, s, s^2, ... , s^{n-1}$$
Proof: To show these elements are linearly independent over $F$, assume $c_0 + c_1s + \cdots + c_{n-1}s^{n-1} = 0$ for some $c_i \in F$.  Let $h(X) = c_0 + c_1 X + \cdots + c_{n-1}X^{n-1}$.  If not all the $c_i$ are zero, then $h$ is a nonzero  polynomial of smaller degree than $f$ which has $s$ as a root, which is impossible.
To show that these elements span $F$, we use the previous theorem that $F(s) = F[s]$: every element of $F(s)$ can be expressed as $h(s)$ for some polynomial $h \in F[X]$.  Using polynomial long division, we can divide $h$ by $f$ to get a quotient and remainder.  That is, $h = qf + r$ for some polynomials $q, r \in F[X]$, where the degree of $r$ is smaller than that of $f$.  Write $r(X) = c_0 + c_1X + \cdots + c_{n-1}X^{n-1}$.  Then $r(s)$ lies in the span of $1, s, ... , s^{n-1}$, with $h(s) = r(s)$.  $\blacksquare$
A: 1) By definition of a field extension:

$$\mathbb{R}(\sqrt{2}) = \{x+y\sqrt{2}:x,y \in \mathbb{R}\}$$

Since $\sqrt{2}$ is already in $\mathbb{R}$, so that $x+y\sqrt{2} \in \mathbb{R}$. Therefore $1$ is the basis for $\mathbb{R}(\sqrt{2})$ over $\mathbb{R}$.
2) Now,

$$\mathbb{Q}(\sqrt[4]{2}) = \{x+y\sqrt[4]{2}:x,y \in \mathbb{Q}\}$$

Here $\sqrt[4]{2} \notin \mathbb{Q}$. The elements $1,\sqrt[4]{2}$ clearly span the $\mathbb{Q}$-vector space $Q(\sqrt[4]{2})$. Now recall that $\sqrt[4]{2} \notin \mathbb{Q}$. If the elements $1,\sqrt[4]{2}$ were linearly dependent we would have $u + v\sqrt[4]{2} = 0$ for some
$u, v \in \mathbb{Q}$ not both zero; in fact it is easy to see that we would then also have $u, v$ both non-zero. Thus we would have 
$$\sqrt[4]{2} = -\frac{u}{v} \in \mathbb{Q}$$
which we know to be false. Hence $1,\sqrt[4]{2}$ are linearly independent and so form a basis for $\mathbb{Q}(\sqrt[4]{2})$ over $\mathbb{Q}$.
A: The answer of D_S is excellent, thorough, and general, but may be a little overwhelming if you're seeing this in the context of linear algebra instead of field theory/Galois theory. Basically, if $F$ is a field and $\alpha\in E$ for some field $E$ for which $F$ is a subfield, think of $F(\alpha)$ as the smallest field containing both $\alpha$ and all of $F$. This essentially boils down to adding in powers of $\alpha$ until you hit a redundant element, and then taking linear combinations over $F$ of $1$ and these non-redundant powers of $\alpha$. (There are technical caveats to this, but addressing them fully gets us more deeply into field theory than we want for the moment. Specifically, we want $\alpha$ to be algebraic over $F$.)
For $\mathbb{R}(\sqrt2)$, we have $\alpha = \sqrt2\in\mathbb{R}$ already, so $\mathbb{R}(\sqrt2)$ is just $\mathbb{R}$ itself. Thus one basis over $\mathbb{R}$ is simply $\{1\}$.
Instead of me doing $\mathbb{Q}(\sqrt[4]2)$ for you, consider the similar example of $\mathbb{Q}(\sqrt[3]5)$. We have $\alpha = \sqrt[3]5\notin\mathbb{Q}$ and $\alpha^2 = \sqrt[3]{25}\notin\mathbb{Q}$, but $\alpha^3 = 5\in\mathbb{Q}$. Then $$\mathbb{Q}(\sqrt[3]5) = \{a+b\sqrt[3]{5}+c\sqrt[3]{25}:a,b,c \in \mathbb{Q}\},$$ which yields a basis over $\mathbb{Q}$ of $\{1, \sqrt[3]{5}, \sqrt[3]{25}\}.$
