Flatness of Residue Field My question refers to a step of in the proof of Corollary 8.5.17 in  Bosch's "Commutative Algebra and Algebraic Geometry";  see page 395
See the red tagged line below:

We consider the exact sequence
$$0 \to I/J \to R[t_1, ..., t_n]/J \to R[t_1, ..., t_n]/I \to 0$$
and we tensor it with $k(s) = \mathcal{O}_{S,s}/m_s$.
Why does it stay exact? Indeed, by assumption $\mathcal{O}_{S,s}$ is flat so it's ok to tensor it with $\mathcal{O}_{S,s}$ but what about $k(s)$? Why does it conserve the exactness?
In addition: Lemma 5.2.9:

 A: Set $T = R[t_1,\ldots, t_n]$. Our assumption that $f$ is flat at $x$ means that $(T/I)_x$ is flat over $R_s$.
We have an exact sequence:
$$ 0 \to I/J \to T/J \to T/I \to 0$$
Localization is exact so we again have the localized exact sequence:
$$0 \to (I/J)_x \to (T/J)_x \to (T/I)_x \to 0$$
Now in 5.2/9, we take $M''$ to be $(T/I)_x$ since it is flat over $R_s$, and take $N$ to be $k(s) = R_s/m_s$. We get the exact sequence we wanted:
$$0 \to (I/J)_x\otimes k(s) \to (T/J)_x\otimes k(s) \to (T/I)_x\otimes k(s) \to 0$$
This justifies the statement "[the sequence] remains exact at $x$ when tensoring over $R$ it with $k(s)$".
(Sweeping it under the rug, but we just used that Localization commutes with tensor products to identify $M\otimes_{R_s} k(s) = M\otimes_{R}k(s)$ for an $R_s$-module $M$ by the way.)
Watch out for the typo in the next line though, $J/I$ should be $I/J$.

Per the title of this question: The residue field $k(s)$ is not going to be a flat module, see e.g. is residue field ever flat over its local ring? on MathOverflow, so it's important that $N$ in 5.2/9 can be any module.
