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A morphism of schemes $f:X\to Y$ is a flat morphism when each local ring ${\mathcal{O}}_{X,x}$ is a flat module over ${\mathcal{O}}_{Y,f(x)}$: tensoring with it preserves exact sequences. An open immersion induces isomorphisms on the local rings at its points, so it is flat.

For a closed immersion, the printed equivalence is valid with the additional hypothesis of finite presentation, for example over a locally Noetherian target, and with nonempty source. Here is the proof with that hypothesis. On an affine chart the map is $A\to A/I$, with $I$ finitely generated. If $\mathfrak{p}\supseteq I$, the flat cyclic module $A_{\mathfrak{p}}/I_{\mathfrak{p}}$ is free over the local ring $A_{\mathfrak{p}}$, and its residue-field dimension is one. Therefore its quotient map from $A_{\mathfrak{p}}$ is an isomorphism and $I_{\mathfrak{p}}=0$. Finite generation gives a neighbourhood of $\mathfrak{p}$ on which $I$ vanishes. The closed image $V(I)$ is consequently open. A nonempty clopen image in a connected target is the whole target; then $I$ vanishes at every prime and is zero. The immersion is an isomorphism. The converse is immediate. This proves the flat closed immersion of finite presentation assertion.

Without finite presentation, connectedness alone does not suffice. Take $A=C([0,1],\mathbb{R})$ and let $\mathfrak{m}=\{a:a(0)=0\}$. Let $I$ consist of functions vanishing on some neighbourhood of zero. The map $A\to A_{\mathfrak{m}}$ has kernel $I$: multiplying such a function by a continuous cutoff equal to one at zero kills it, whereas $ga=0$ with $g(0)\ne 0$ forces $a$ to vanish near zero. This map is surjective, because the reciprocal of any denominator nonzero at zero can be extended continuously from a small neighbourhood of zero to the whole interval. ThusThe map $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$ is both a closed immersion and a flat localization. Its source is connected because its ring is local. Its target is connected because continuous idempotent functions on the connected interval can only be the constants zero and one. It is not an isomorphism: $I$ contains nonzero functions supported away from zero. This supplies a concrete counterexample to the unrestricted assertion. The kernel is a pure ideal; this is the general mechanism behind such examples, as explained in the Stacks Project's treatment of pure ideals.

For the finite-flat fibre assertion, first note the local algebra fact that a finitely generated flat module $M$ over a local ring $(A,\mathfrak{m})$ is free. Choose generators $m_1,\dots ,m_r$ whose residues form a basis of $M/\mathfrak{m}M$, using Nakayama's lemma. If a row $a=(a_1,\dots ,a_r)$ gives a relation $\sum a_i{m}_i=0$, the equational criterion for flatness supplies elements $n_j$ and a matrix $B=(b_{ij})$ such that $m_i={\sum }_j{b}_{ij}{n}_j$ and $aB=0$. Write each $n_j$ as a linear combination of the $m_i$, with coefficient matrix $C$. Then $m=BCm$. Independence modulo $\mathfrak{m}$ makes $BC$ congruent to the identity. This square matrix is invertible over the local ring, and $aBC=0$ implies $a=0$. Thus the chosen generators are a basis. This also justifies the cyclic-module step above; see the finite-flat local lemma.

Now let $\mathcal{A}=f_{\ast }{\mathcal{O}}_X$ for the finite flat morphism. On each affine chart this is the finite flat algebra defining its inverse image. Hence ${\mathcal{A}}_y$ is free over ${\mathcal{O}}_{Y,y}$, of some finite rank $r_y$. Let $\eta$ be the generic point of the irreducible target. Localizing that free module at $\eta$ givesEvery point specializes from the same $\eta$, so every $r_y$ equals the fixed rank at $\eta$. This remains valid if the irreducible target is nonreduced: one compares free ranks over its nonzero generic local ring, not necessarily over a field. The fibre is affine, andConsequentlyThis is finite flat rank over an irreducible scheme. No unjustified replacement of finite flat by finite locally free on an arbitrary base is needed.

For an algebraic curve of finite type over a field, its normalization of an integral scheme is finite and birational. If flat, the preceding rank is one, since the generic fibre has the same function field. Locally its normalization algebra $B$ is therefore a free $A$-module of rank one. The element $1_B$ generates: its image cannot vanish in the one-dimensional fibre algebra, and Nakayama's lemma applies. Thus $A\to B$ is an isomorphism. Conversely the identity normalization is flat.

The parenthetical definition in the PDF does not explicitly impose finite type. One can prove the conclusion even in that broader setting without assuming a finite normalization. For any affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ describe its normalization of an integral scheme. Integrality makes the map on spectra surjective; if $B$ is flat it is faithfully flat. Faithful flatness gives contraction of ideals, $vB\cap A=vA$. For $b=u/v\in B$, with $u,v\in A$ and $v\ne 0$, one has $u=vb\in vB\cap A=vA$. Thus $u=va$ for some $a\in A$, so $b=a$. Therefore $B=A$ on every chart. We have proved the flat normalization is an isomorphism result and, in particular,