A field extension is a triple $(k,\tau,L)$ where $\tau:k \ra L$ is a non zero ring homomorphism. $L$ is called the extension field of $k$ and $\tau$ is called an embedding of $k$ into $L$. The subfield $\tau(k)$ of $L$ is often identified with $k$, since $\tau(k)$ is isomorphic to $k$ ,and the extension is written as $L/k$.
 For a fixed $k$ and $L$ there can be many different field embeddings, for example for $k=\Q(\ci)$ and $L=\C$, $\tau$ could be identity map or a conjugation.

Consider a field extension $L/k$ and $S$ a subset of $L$. The intersection of all fields of $L$ which contain both $k$ and $S$ is called the subfield of $L$ generated by $k$ and $S$. It is denoted by $k(S)$ and is the smallest subfield containing both $k$ and $S$. If $S=\{\alpha_1,\ldots,\alpha_n\}$, then $k(S)$ is written as $k(\alpha_1,\ldots,\alpha_n)$ (also see compositum ).
 The definition rests on the fact that if $\{F_i\}_{i\in I}$ is a collection of subfields of $L$, then $\cap_iF_i$ is also a subfield of $L$.

A field extension $L/k$ is finitely generated if $L=k(\alpha_1,\ldots,\alpha_n)$ for a set $S=\{\alpha_1,\ldots,\alpha_n\}\in L$. In case $L=k(\alpha)$ (a single element is adjoined) the extension is called simple.

$L$ is a $k$ vector space for a field extension $L/k$ . The dimension of the vector space is called degree of extension denoted as $[L:k]$. If degree is finite $L/k$ is called a finite extension.

$[\C:\RR]=2$ with basis $\{1,\ci\}$ or $\{2-5\ci, 7+\ci\}$
$[\Q(\sqrt{2}):\Q]=2$ with basis $\{1,\sqrt{2}\}$
$[\C:\Q]=\infty$ also $[\RR:\Q]=\infty$
$[k(X):k]=\infty$

Consider two field extensions $L/k$ and $K/k$, where both $L$ and $K$ sit inside a bigger field $\Omega$. Then the compositum $KL$ is the intersection of all subfields of $\Omega$ that contain both $L$ and $K$. More formally, $KL=k(K\cup L)=K(L)=L(K)$.
  The map $K\otimes_k L\ra KL$ is injective, and when degrees are finite it becomes bijective.

Proof 1 Let $(a_1,\ldots, a_n)$ be a $k$ basis for $K/k$, then the following is the finite sequence of extensions. $$K\subset K(a_1)\subset K(a_1,a_2)\subset\ldots\subset KL $$

Proof 2 Since the extensions are finite the map $K\otimes_k L\ra KL$ is bijective, since $K\otimes_k L$ is finite dimensional so is $KL$.

Given $L/k$ a field extension and $\alpha\in L$, consider the evaluation map $\varphi_{\alpha}:k[X]\ra L$ where $f(X)\mapsto f(\alpha)$, the kernel of this map is an ideal generated by all polynomials $f(X)\in k[X]$ that have $\alpha$ as a root $$\ker\varphi_{\alpha}=\id{m}_\alpha=\{f\in k[X]\text{ such that }f(\alpha)=0\}. $$ If $\id{m}_\alpha\neq 0$ then $\alpha$ is algebraic over $k$, else $\alpha $ is transcendental over $k$. The field $L$ is algebraic over $k$ if every element of $L$ is algebraic over $k$. Note that $k$ is algebraic over $k$ by considering the ideal $(X-a), a\in k$.
  Since $k$ is a field $k[X]$ is a principal ideal domain, thus $\id{m}_\alpha$ is generated by a single polynomial say $m_\alpha(X)$ of minimal degree in $\id{m}_\alpha$. This polynomial is called the minimal polynomial of $\alpha$ over $k$. Note that $m_\alpha(X)$ is an irreducible polynomial of least degree in $\id{m}_\alpha$, because if it factorizes non trivially into two polynomials then those polynomials would have lower degree than $m_\alpha(X)$ contradicting our choice of minimal degree in $\id{m}_\alpha$. Furthermore, for the same reason $\id{m}_\alpha$ is also maximal, as division is containment and non trivial division would require a polynomial of a lower degree.

  A complex number $\alpha\in\C$ is algebraic if there a non zero polynomial $f(X)\in\Q[X]$ such that $f(\alpha)=0$ else, $\alpha$ is transcendental.

$\RR/\Q$, $\alpha=\sqrt[3]{2}$ is algebraic with minimal polynomial $X^3-2$
$\C/\RR, \alpha=\sqrt[3]{2}$ is algebraic with minimal polynomial $X-\sqrt[3]{2}$
$\C/\Q$ then for every integer $n\geq 1$ and $a\in\Q$, $\alpha=\sqrt[n]{a}$ is algebraic and minimal polynomial is $X^n-a$.
$\pi$ is transcendental.

Proof First note that $\lr{f(X)}$ is maximal, thus $k[X]/\lr{f(X)}$ is a field and let $\wbr{X}$ be the image of $X$ under the mapping $$k[X]\ra k[X]/{\lr{f(X)}}\qquad X\mapsto\wbr{X}=X+\lr{f(X)}$$ Note that $f(\wbr{X})=f(X+\lr{f(X)})\in\lr{f(X)}$ in $E$ by construction. Thus, $\wbr{X}$ is the root of the polynomial $f(X)$ in the field $E$. The basis of $E/k$ is given by $\{1,\wbr{X},\ldots,\wbr{X}^{n-1}\}$, and since the ideal $\lr{f(X)}$ does not contain any polynomials of degree less than $n$, this basis is independent.

$E=\Q[X]/\lr{X^3-2}$ is isomorphic to three fields $\Q(\sqrt[3]{2}),\Q(\sqrt[3]{2}\omega),\Q(\sqrt[3]{2}\omega^2)$ where $\omega$ is the third root of unity, these are three different embeddings of $E$ into a larger field $\C$. Thus, $\wbr{X}$ has three different realizations. Notice that, $\Q(\sqrt[3]{2}\omega)\subset\C$ whereas $\Q(\sqrt[3]{2})\subset \RR$.

(a)  Let $\alpha\in L$ and $[L:k]=n$, then we can construct a linearly dependent set of $n+1$ elements $\{1,\alpha,\alpha^2,\ldots,\alpha^n\}$ such that $$a_0\alpha^n+a_1\alpha^{n-1}+\ldots+a_{n-1}\alpha+a_0=0,\qquad a_i\in k $$ and not all $a_i$ are zero in $k$. Replacing $\alpha$ by $X$ gives a non trivial polynomial with the root $\alpha$. Thus, $\alpha$ is algebraic over $k$.
(b)  If $\alpha$ is algebraic with a minimal polynomial, then by proposition above $[E:k]=$ degree of minimal polynomial (and thus finite).
If $E/k$ is finite then by part (a) setting $L=E$ we get $E$ is algebraic over $k$, since $\alpha\in E$, it is algebraic over $k$.

Proof If one of the extensions $F/L$ or $L/k$ is infinite then this is just $\infty=\infty$. If both are finite extensions let $\{u_i\}_{i\in I}$ be the basis of $F/L$ and $\{v_j\}_{j\in J}$ be the basis of $L/k$. Then the basis of $F/k$ is given as $\{u_iv_j\}_{i\in I,j\in J}$.

Proof by induction Since, $\alpha_1$ is algebraic by lemma(b) above $k(\alpha_1)/k$ is finite. Induction hypothesis $\alpha_{i+1}$ is algebraic over $K:=k(\alpha_1,\ldots,\alpha_i)$ thus $K(\alpha_{i+1})/K$ is finite. Now, by tower law $k(\alpha_1,\ldots,\alpha_{i+1})/k$ is finite or each extension below is finite and hence ther result. $$ [k(\alpha_1,\ldots,\alpha_{n-1},\alpha_n):k(\alpha_1,\ldots,\alpha_{n-1})]\cdots[k(\alpha_1,\alpha_2):k(\alpha_1)][k(\alpha_1):k]$$

(a) $[\Q(\sqrt{2}): \Q] = 2$ since, its minimal irreducible polynomial is $X^2-2$ in $\Q[X]$ and $\{1, \sqrt{2}\}$ is a basis of $\Q(\sqrt{2})/\Q$.
(b) $[\Q(\sqrt[3]{2}):\Q]=3$ since, its minimal irreducible polynomial (by Eisentein Criterion) is $X^3 - 2$ in $\Q[X]$.
(c) The above two examples imply that $\sqrt[3]{2}\not\in \Q(\sqrt{2})$, Otherwise by tower law $$[\Q(\sqrt{2}):\Q] = [\Q(\sqrt{2}):\Q(\sqrt[3]{2})][\Q(\sqrt[3]{2}):\Q].$$ which would not hold since LHS is $2$ and RHS has a $3$ as a factor. Similary, $\sqrt{2}\not \in \Q(\sqrt[3]{2})$, else $$[\Q(\sqrt[3]{2}):\Q] = [\Q(\sqrt[3]{2}):\Q(\sqrt{2})][\Q(\sqrt{2}):\Q].$$ which would not hold since LHS is $3$ and RHS has a $2$ as a factor.
(d) Let us compute $[\Q(\sqrt[3]{2},\sqrt{2}):\Q]$ $$ \begin{align*} [\Q(\sqrt[3]{2},\sqrt{2}):\Q] &= [\Q(\sqrt[3]{2},\sqrt{2}):\Q(\sqrt{2})][\Q(\sqrt{2}):\Q]\\ &=[\Q(\sqrt[3]{2},\sqrt{2}):\Q(\sqrt{2})]\cdot 2\\ &= [\Q(\sqrt[3]{2},\sqrt{2}):\Q(\sqrt[3]{2})][\Q(\sqrt[3]{2}):\Q]\\ &= [\Q(\sqrt[3]{2},\sqrt{2}):\Q(\sqrt[3]{2})]\cdot 3\\ \end{align*} $$ The above shows that $[\Q(\sqrt[3]{2},\sqrt{2}):\Q]$ is divisible by $6$ (that is both $2$ and $3$), thus degree is atleast $6$. On the other hand $X^2-2\in\Q[X]$ has degree $2$ with root $\sqrt{2}$ and $X^3-2\in \Q(\sqrt{2})[X]$ has degree $3$ with root $\sqrt[3]{2}$, thus the extension $[\Q(\sqrt[3]{2},\sqrt{2}):\Q] $ has degree atmost $6$ (coming from the two irreducible polynomials). Hence, $[\Q(\sqrt[3]{2},\sqrt{2}):\Q]=6$