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  "body": "\n## Definition\n\nLet $A$ be a commutative ring and let $B$ be a commutative [[$A$-algebra|DPBJ04JMH8]].\nThe algebra $B$ is *finitely presented* over $A$ if it has a presentation with finitely many generators and relations.\nThat means there are an integer $n \\ge 0$ and a finitely generated ideal $I = (g_1, \\ldots, g_m)$ of the [[polynomial ring|0SDC934R52]] $A[x_1, \\ldots, x_n]$ with $B \\cong A[x_1, \\ldots, x_n]/I$ as $A$-algebras.\nHere $A[x_1, \\ldots, x_n]/I$ is the [[quotient ring|0076HWHGHQ]].\n\n## Remarks\n\nExample: $\\Z[i]$ is finitely presented over $\\Z$, since $\\Z[i] \\cong \\Z[x]/(x^2 + 1)$ and the ideal $(x^2 + 1)$ is generated by one element.\n\nNon-example: let $A = k[y_1, y_2, \\ldots]$ be a polynomial ring over a field $k$ in countably many variables, and let $B = A/(y_1, y_2, \\ldots) \\cong k$.\nThen $B$ is of [[finite type|AC5TFFQT9M]] over $A$, generated by no elements at all, but it is not finitely presented.\nIf an algebra is finitely presented, then the kernel of every surjection $A[x_1, \\ldots, x_n] \\to B$ is finitely generated, and the kernel of $A \\to B$ is $(y_1, y_2, \\ldots)$, which is not.\n\nIf $A$ is Noetherian, every finite type $A$-algebra is finitely presented, because every ideal of $A[x_1, \\ldots, x_n]$ is finitely generated by [[Hilbert's basis theorem|0YKA9J3XEE]].\n",
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