Finitely presented algebra

Remarks

Example: $\Z[i]$ 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.

Non-example: let $A = k[y_1, y_2, \ldots]$ be a polynomial ring over a field $k$ in countably many variables, and let $B = A/(y_1, y_2, \ldots) \cong k$. Then $B$ is of finite type over $A$, generated by no elements at all, but it is not finitely presented. If 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.

If $A$ 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.