Algebraic and Transcendental Numbers
Let $K$ be a field and $E/K$ be a field extension. We call an element $\alpha\in E$ algebraic over $K$ if there exists a nonzero polynomial $f(x)\in K[x]$ such that $f(\alpha) = 0$. If no such polynomial exists we say that $\alpha$ is transcendental over $K$ if there does not exist such polynomial. It is easy to see that any element $\alpha\in K$ is algebraic over $K$ since it is the root of the polynomial $f(x) = x - \alpha$.
To understand this definition better we can let $K=\mathbb{Q}$ and $E=\mathbb{C}$, then $\sqrt{2}$, $\sqrt[5]{541}$, $i$ are all algebraic over $\mathbb{Q}$. Numbers in $\mathbb{C}$ which are algebraic over $\mathbb{Q}$ are called algebraic numbers, and which are transcendental over $\mathbb{Q}$ are called transcendental numbers. We will later give sufficient conditions for a number to be transcendental (or necessary conditions for a number to be algebraic) and construct a number that is transcendental.
If $\alpha\in E$ is algebraic over $K$ let $I=\{g(x)\in K[x]: g(\alpha) = 0 \}$. It is straightforward to check that this is an ideal of $K[x]$. Then since $K$ is a field, $K[x]$ is a principal ideal domain (even a Euclidean domain1), so $I=f(x)K[x]$ for some $f(x)\in K[x]$. We can see that any generator of $I$ has the minimal degree such that any other polynomial in $K[x]$ that has a smaller degree will not be in $I$. We can let $f(x)$ be monic, since if otherwise we can multiply it by its leading coefficient’s inverse and the result will again be a generator of $I$. Then $f(x)$ is unique, since if another generator is monic, then it must be $1$ times $f(x)$. We call this unique polynomial the minimal polynomial of $\alpha$ over $K$. The degree of the minimal polynomial is called the degree of $\alpha$ over $K$. We see that $\alpha\in K$ if and only if $\alpha$ is degree $1$ over $K$.
We can look at examples from algebraic numbers. $\sqrt{2}$ is degree $2$ over $\mathbb{Q}$. Its minimal polynomial is $x^2-2$. It is easy to prove this because we can prove that $\sqrt{2}$ is not a rational number so its minimal polynomial is at least degree $2$. And since we can find a polynomial of degree $2$, then its degree must be $2$. Similar arguments show that $\sqrt{3}$ and $i$ are also degree $2$ over rationals.
Now notice that, $f(x)\in K[x]$ is the minimal polynomial of $\alpha\in E$ if and only if $f(\alpha) = 0$, $f(x)$ is monic and $f(x)$ is irreducible in $K[x]$. So if we can show that a monic polynomial of degree $n$ is irreducible then we know that all of the roots of this polynomial has its minimal polynomial this polynomial.
For any $n\in\mathbb{Z}^+$ and $p$, $x^n-p$ is irreducible by Eisenstein’s Criterion. Then we have that $\sqrt[n]p$ is of degree $n$ over the rationals for any $n$ and any $p$ prime. You can also easily generalize this by using the full potential of Eisenstein’s Criterion, but I will not do that. For example, this gives $\sqrt[5]{4079}$ is degree $5$ over rationals.
A degree $2$ or $3$ rational polynomial is irreducible if and only if it does not have a rational root.
A non-constant polynomial with integer coefficients is irreducible in $\mathbb{Z}[x]$ if and only if it is primitive and irreducible in $\mathbb{Q}[X]$ by Gauss’s Lemma. Then if $f(x)$ is a degree $2$ primitive polynomial with integer coefficients, $f(x)$ is irreducible if and only if it does not have any integer roots. $x^2+x+1$ is irreducible because it does not have any integer roots, and to see this you can just examine this polynomial in $\mod 2$, (since $x^2+x\equiv 0 \mod 2$). Of course we can just check the roots of a degree $2$ polynomial to see if it is irreducible, but this method might provide a fast solution. The real power of this method though, is when $f(x)$ is a degree $3$ primitive integer polynomial. It is still correct that $f(x)$ is irreducible if an only if it does not have an integer root. Then for example we can prove that $2x^3+x^2+x+1$ is irreducible since it does not have any integer roots (again, examine it in $\mod 2$).
Adjoining Algebraic Elements
Let $C$ be the collection of all subfields of $E$ that contains $K$ and $\alpha$. Then we define $K(\alpha)$ as the smallest subfield of $E$ in $C$, that is $K(\alpha) = \bigcap C$. $K(\alpha)$ is a field extension of $K$.
Another definition of $K(\alpha)$ is as the field of fractions of $K[\alpha]$, it is easy to prove that these two definition are equivalent.
If $\alpha$ is algebraic over $K$, then the degree of $K(\alpha)$ over $K$ is the degree of $\alpha$ over $K$. So $\alpha$ is algebraic over $K$ if and only if $K(\alpha)$ is a finite extension. Or, in other words, $\alpha$ is algebraic over $K$ if and only if it is contained in a finite extension of $K$, and its degree over $K$ is at most the degree of that extension over $K$.
Now let’s prove this. $f(x)K[x]$, as we have already discussed, is an ideal of $K[x]$. Thus $K[x]/f(x)K[x]$ is also a ring, and since it contains $K$, it is also a vector space over $K$. Its dimension is at least $n=\deg(f(x))$, since $1,x,\dots,x^{n-1}$ is linearly independent (since any linear combination of them has degree less than $f(x)$, they can not form a nonzero multiple of $f(x)$). We will show that in fact $1,x,\dots,x^{n-1}$ spans $K[x]/f(x)K[x]$. Now clearly the set $1,x,x^2,\dots$ spans this set, so to show that $1,x,\dots,x^{n-1}$ also spans, it is enough to show that they span $1,x,x^2,\dots$. First of all, they clearly span $x^m$ for $n<m$. Now we show that they span $x^n$. Notice $x^n=x^n-f(x)$, and $x^n-f(x)$ has degree less than $n$, so it can be spanned. Now to prove by induction, assume they span $x^i$ for all $i<m$ for some $m>n$, then we will show that they also span $x^m$. Notice $x^m=x^m-x^{m-n}f(x)$, and $x^m-x^{m-n}f(x)$ has degree less than $m$, so they must span it by the induction hypothesis, and so we are done.
$f(x)K[x]$ is a maximal ideal of $K[x]$, since $f(x)$ is irreducible. Then $K[x]/f(x)K[x]$ is a field since its ideals are only the trivial ideals. Now, let $\varphi:K[x]\to K(\alpha)$ be the evaluation homomorphism $\varphi(h(x)) = h(\alpha)$. Then clearly the kernel of this homomorphism is $f(x)K[x]$, thus $K[x]/f(x)K[x]$ is isomorphic to the image of $\varphi$, and the image of $\varphi$ is then a field. The image contains $K$, which can be seen by applying the evaluation homomorphism to the constant polynomials. It also contains $\alpha$, since $\varphi(x)=\alpha$. Then the image must be $K(\alpha)$, by the very definition of $K(\alpha)$ being the smallest field containing both $K$ and $\alpha$. Thus $K[x]/f(x)K[x]$ is isomorphic to $K(\alpha)$, and since they both contain $K$, this is a vector space isomorphism over $K$. So their dimensions are equal, so the dimension of $K(\alpha)$ over $K$ is $n=\deg(f(x))$ which is the degree of $\alpha$ over $K$.
$K(\alpha)(\beta)=K(\beta)(\alpha)$. This can easily be seen by the first definition. Then we will denote $K(\alpha)(\beta)=K(\alpha,\beta)$.
Observe that if $\alpha$ and $\beta$ are algebraic over $K$, with degrees $n,m$ respectively, then $K(\alpha,\beta)$ is at most a degree $nm$ extension, since $\beta$ is at most an $m$ degree extension over $K(\alpha)$. Then we see that $\alpha+\beta,\alpha\beta,\alpha/\beta$ are all algebraic over $K$, since they are all elements of a finite extension, namely $K(\alpha,\beta)$. So that all algebraic elements over a field themselves form a field. This field will be called the algebraic closure of $K$, and will be denoted by $\overline K$.
$\sqrt2+\sqrt[5]3$ is then an algebraic number.
Let $\alpha$ be degree $n$ over $K$ and $\beta$ be degree $m$ over $K$. If $n$ and $m$ are relatively prime, then the degree of $K(\alpha,\beta)$ over $K$ is $nm$. This is because both $n$ and $m$ must divide the degree of $K(\alpha,\beta)$ over $K$, as $K(\alpha,\beta)$ is a field extension of both $K(\alpha)$ and $K(\beta)$. If we assume $K$ is of characteristic $0$, more is true: $\alpha+\beta$ is degree $nm$ over $K$, too.2 With this fact, it is trivial that $\sqrt2+\sqrt[5]3$ is of degree $10$ over the rationals.
Let’s also give an elementary proof. Similarly, we will show that $\mathbb{Q}(\sqrt2+\sqrt[5]3)=\mathbb{Q}(\sqrt2,\sqrt[5]3)$. The minimal polynomial of $\sqrt2$ is $f(x)=x^2-2$ over $\mathbb{Q}$. Then $f(\sqrt2+\sqrt[5]3-x)$ is a polynomial in $\mathbb{Q}(\sqrt2+\sqrt[5]3)$ and one of its roots is $\sqrt[5]3$. As can be checked, the other root of this polynomial is not a fifth root of $3$, so the only root $f(\sqrt2+\sqrt[5]3-x)$ shares with $x^5-3$ is $\sqrt[5]3$. Then their greatest common divisor in $\mathbb{Q}(\sqrt2+\sqrt[5]3)$ is $(f(\sqrt2+\sqrt[5]3-x),x^5-3)=x-\sqrt[5]3$, and so $x-\sqrt[5]3\in \mathbb{Q}(\sqrt2+\sqrt[5]3)[x]$, thus $\sqrt[5]3\in \mathbb{Q}(\sqrt2+\sqrt[5]3)$. Then $\sqrt2+\sqrt[5]3-\sqrt[5]3=\sqrt2\in\mathbb{Q}(\sqrt2+\sqrt[5]3)$, thus $\mathbb{Q}(\sqrt2+\sqrt[5]3)=\mathbb{Q}(\sqrt2,\sqrt[5]3)$.
$\overline{\mathbb{Q}} \subseteq \mathbb{C}$, since every rational polynomial can be factored into degree $1$ factors in $\mathbb{C}$, by the fundemental theorem of algebra.
Liouville’s Theorem
We have only seen algebraic numbers so far. Perhaps $\overline{\mathbb{Q}} = \mathbb{C}$, and thus there are no transcendental numbers? This is not true as we will see now.
Liouvielle’s theorem says that if $a\in\mathbb{R}$ is an algebraic number of degree $n>1$ (so that it is irrational), then there exists a number $c$ depending only on $a$, such that $|a-\frac pq| > \frac c{q^n}$ for all rationals $\frac pq$, where $q>0$. So that, to prove a number $a$ is transcendental, it is sufficient to show that for any $c$ and $n$ there exists a rational $\frac pq$ such that $|a-\frac pq| \leq \frac c{q^n}$.