A finite flat module over a Noetherian ring is finite projective

By [L1], choose a finite presentation $R^m\stackrel{D}{\to }{R}^n\stackrel{\pi }{\to }M\to 0$. Put $x_i=\pi (e_i)$, and write the columns of $D$ as relation rows $u^{(1)},\dots ,u^{(m)}\in R^n$, so $\ker \pi$ is generated by those rows. Start with $x=Ay$, where $A$ is the identity matrix and $y$ is the column of the $x_i$.

Process the $m$ relation rows in order. Suppose $x=Ay$ and the first $h-1$ rows satisfy $u^{(j)}A=0$. For the next row $u^{(h)}$, the equality $u^{(h)}x=0$ says $(u^{(h)}A)y=0$. Applying [L2] to this one finite relation gives a finite column $z$ of elements of $M$ and a matrix $B$ with $y=Bz$ and $(u^{(h)}A)B=0$. Replace $(A,y)$ by $(AB,z)$. The new matrix satisfies $u^{(h)}AB=0$, and each earlier relation remains zero because $u^{(j)}AB=(u^{(j)}A)B=0$. After the finite last stage, write $x=Cz$ with $u^{(j)}C=0$ for every $j$. This finite induction makes only finitely many choices.

Let $z$ have $t$ entries, and define $s_0:R^n\to R^t$ by sending $e_i$ to the $i$th row of $C$. Every generator $u^{(j)}$ of $\ker \pi$ is killed by $s_0$ because $u^{(j)}C=0$. Thus $s_0$ descends to $s:M\to R^t$. Define $q:R^t\to M$ by sending its basis vectors to the entries of $z$. The identity $x=Cz$ says $qs(x_i)=x_i$ for every generator $x_i$, hence $qs=1_M$.

Let $E\twoheadrightarrow N$ be any epimorphism and $g:M\to N$ any map. For each of the finitely many basis vectors of $R^t$, choose a preimage in $E$ of its image under $gq$, and extend linearly to a lift $h:R^t\to E$. Then $hs$ lifts $g$ because $qs=1_M$. By [L3], $M$ is projective. It is finite by hypothesis, so it is finite projective.