An algebra A over a field k is finitely generated when there are finitely many elements a_1, \ldots, a_n \in A such that every element of A is a polynomial in them, written A = k[a_1, \ldots, a_n]. The a_i are the generators, and only addition, subtraction and multiplication are allowed: no division.
The definition can also be read as a surjection. Sending x_i \mapsto a_i gives a surjection k[x_1, \ldots, x_n] \twoheadrightarrow A from the polynomial ring in n variables, and with kernel I we get A \cong k[x_1, \ldots, x_n]/I. Every finitely generated k-algebra has this form.
The generators are not unique, and even their number can change. In the figure every monomial of total degree up to d is a product of the two generators x and y, and as d grows the monomials fill the whole plane of exponents.
Being finitely generated as a k-algebra and being finite as a k-module are different conditions. The ring k[x] is generated by the single element x, yet as a vector space over k it is infinite-dimensional, since 1, x, x^2, \ldots are linearly independent.
Finite is the stronger condition: a k-algebra that is finite-dimensional over k is certainly finitely generated. The converse fails, and k[x] is the counterexample, while k[x]/(x^2 - 1) satisfies both, being two-dimensional over k and generated by x.
The figure nests the three classes: all k-algebras outside, finitely generated ones in the middle and finite-dimensional ones inside, each with examples that belong to that layer only. Algebraic geometry uses the middle layer as coordinate rings, and the outer layer appears on the side, as completions and function fields.
The polynomial ring k[x_1, \ldots, x_n] is the definition itself, with the n variables as generators. The coordinate ring of the parabola, k[x, y]/(y - x^2) \cong k[t], can be generated by two elements, but t alone is enough.
The coordinate ring of the hyperbola is k[x, y]/(xy - 1) \cong k[t, t^{-1}]. Here t alone does not suffice, because t^{-1} cannot be made from it: the generators are t and t^{-1}. Likewise the localization at one element is k[x]_f = k[x, 1/f], generated by x and 1/f.
These last two allow exactly one denominator. Adding that denominator as a new generator keeps everything inside the world of polynomials. When denominators are allowed without restriction, the trick no longer works.
The rational function field k(x) is not finitely generated as a k-algebra [1]. Suppose k(x) = k[f_1/g_1, \ldots, f_n/g_n] and put g = g_1 \cdots g_n. Every polynomial in the generators has a power of g as denominator, so every element of k(x) would have the form h/g^m.
The ring k[x] has infinitely many irreducible polynomials, so we can choose one, p, that does not divide g. If 1/p = h/g^m, then g^m = ph. Since p is irreducible it is prime, and dividing g^m it must divide g, a contradiction.
The argument works even when k is finite, because it only uses that k[x] has infinitely many irreducible polynomials. In the figure the first dots are the irreducible factors of g and can be used as denominators; all the others, continuing without end, cannot.
A subring of a finitely generated algebra need not be finitely generated [1]. In k[x, y] collect the polynomials that are constant along the y-axis, those with f(0, y) independent of y. They form a ring A = k[x, xy, xy^2, xy^3, \ldots].
Suppose finitely many generators were enough, and let N bound the y-degrees of their parts of degree 1 in x. Multiplying generators raises the x-degree, so the part of degree 1 in x of any product is a combination of those parts, and its y-degree stays at most N. Then xy^{N+1} lies in A but cannot be produced.
Counting gives another example. Over k = \mathbb{Q} a finitely generated algebra is a quotient of a polynomial ring and so is countable, while \mathbb{Q}[[t]] corresponds to all sequences of coefficients and is uncountable. The formal power series ring k[[t]] is therefore not finitely generated.
Let A = k[a_1, \ldots, a_n] and let I be an ideal. Then A/I is generated by the images \bar{a}_1, \ldots, \bar{a}_n, so it is finitely generated. The localization A_f at one element f is k[a_1, \ldots, a_n, 1/f]: one more generator.
Localizing at a prime ideal is different. A_{\mathfrak{p}} allows every element outside \mathfrak{p} as a denominator, and in general it is not finitely generated. In k[x]_{(x)} every x - a with a \neq 0 may appear in a denominator, and the argument for the rational function field goes through again.
Field extensions use the words in another sense. K/k is finitely generated as a field extension when K = k(\alpha_1, \ldots, \alpha_n), with fractions allowed. So k(x) is finitely generated as a field, by x alone, but not as a k-algebra: the difference is in the brackets, k[\ldots] against k(\ldots).
By Hilbert's basis theorem, if k is Noetherian so is k[x_1, \ldots, x_n], and quotients of Noetherian rings are Noetherian. So every finitely generated k-algebra is Noetherian: every ideal is generated by finitely many elements, and every figure is cut out by finitely many equations.
There are also only finitely many minimal primes, so a figure has finitely many irreducible components. The converse fails: k[[t]] is Noetherian but not finitely generated.
The Krull dimension is finite too, by Noether normalization [3]: there are algebraically independent y_1, \ldots, y_d \in A with A finite over k[y_1, \ldots, y_d], and this d is exactly \dim A. For a domain it equals the transcendence degree of the fraction field [1], and \dim A/\mathfrak{p} + \operatorname{ht} \mathfrak{p} = \dim A holds for every prime \mathfrak{p} [4].
If a field K is finitely generated as an algebra over a field k, then K is a finite extension of k [2]. This is Zariski's lemma, and it needs no assumption on k. The failure of k(x) is a special case: k(x) is not finite over k, so it cannot be finitely generated.
For uncountable k a dimension count proves it [1]. A finitely generated L = k[e_1, \ldots, e_n] is spanned by the monomials in the generators, which are countably many, so L has countable dimension over k. The field k(T) has uncountable dimension, since the functions \dfrac{1}{T - \lambda} for \lambda \in k are linearly independent.
If the field L contained an element t transcendental over k, then T \mapsto t would embed k(T) in L and give L uncountable dimension. So every element of L is algebraic over k, and a ring generated by finitely many algebraic elements is finite over k. Over \mathbb{C} this settles it; general fields need another route [2].
Let A be finitely generated over k and \mathfrak{m} a maximal ideal. Then A/\mathfrak{m} is a field finitely generated over k, and by Zariski's lemma it is a finite extension of k. When k is algebraically closed the only finite extension is k itself, so A/\mathfrak{m} = k.
For A = k[x_1, \ldots, x_n] the image of each x_i is some a_i \in k, so every x_i - a_i lies in \mathfrak{m} and (x_1 - a_1, \ldots, x_n - a_n) \subseteq \mathfrak{m}. The left side is already maximal, so the two are equal [2]. This is why maximal ideals may be called points.
Without algebraic closure it breaks down. In \mathbb{R}[x] the ideal (x^2 + 1) is maximal, but its residue field is \mathbb{C} and it corresponds to no real point [2]: the graph of x^2 + 1 never meets the axis.
The number of generators depends on the ring. k[t] needs one; k[t^2, t^3] needs two, since t^2 alone cannot make t^3 and t^3 alone cannot make t^2; k[t, t^{-1}] needs two; k[x_1, \ldots, x_n] needs n; and for k(x) no finite number is enough.
The ring k[t^2, t^3] is the subring of k[t] without t, the coordinate ring of the cusp. The smallest number of generators is the dimension of the smallest affine space containing the figure: the cusp fits in the plane but not on a line.
This is the two-way road between rings and figures. A finitely generated domain over k is a quotient of a polynomial ring, and the zeros of its ideal give a figure; conversely the coordinate ring of an irreducible affine variety is a finitely generated domain. Rings that are not finitely generated have no classical figure behind them.