Example of a full set of sections on product scheme that is not one on each summand In Katz and Mazur's book "Arithmetic Moduli of Elliptic Curves" (available here), the following statement is made about full set of sections and fibered product (Lemma 1.8.5, page 35).  

Let $Z_1/S$ and $Z_2/S$ be two non-empty finite flat $S$-schemes of finite presentation and ranks $N_1$, $N_2$ respectively. Let 
  $$P_1^{(1)},\ldots ,P_{N_1}^{(1)}\in Z_1(S)$$
  $$P_1^{(2)},\ldots ,P_{N_2}^{(2)}\in Z_2(S)$$
  be given sequences of $N_i$ points in $Z_i(S)$ for $i=1,2$. Suppose that $Z_1/S$ is finite étale. Then the following conditions are equivalent:
   1. For $i=1,2$, the $N_i$ points $P_1^{(i)},\ldots ,P_{N_2}^{(i)}$ in $Z_i(S)$ form a full set of sections of $Z_i(S)$.
   2. The $N_1N_2$ $S$-valued points $P_j^{(1)}\times P_k^{(2)}$ of $Z_1\times _S Z_2$ form of full set of sections of $Z_1\times _S Z_2$.

(For the definition of a full set of sections, please look at the book or read this related question).  
After the proof of this statement, it is stated that if we drop the hypothesis that $Z_1$ is étale over $S$, then we only have $1. \Rightarrow 2.$  
I have been trying to find a counterexample to disprove $2. \Rightarrow 1.$, but I have been unable to find one. I tried to look at cases where $S=\operatorname{Spec}(k)$ is the spectrum of a field, and $Z_1$, $Z_2$ the spectra of non-reduced $k$-algebras (so that they are not étale), such as $k[\epsilon]/(\epsilon^2)$, but I couldn't find anything satisfying. Sadly, my intuition about étaleness is not developped enough to have a feeling of what could work here.  
Would someone here please be able to provide an example?
I thank you very much for your help.
 A: I noticed later that in fact, two examples are given in (1.11.4) and (1.11.5), pages 52-55 in the book. I am going to write the second one down as an answer to my own question.  

Let $p$ be a prime, and $S=\operatorname{Spec} (\mathbb{Z}/p^2\mathbb{Z})$. We let $Z_1=Z_2= \mu_p$ be the $p$-roots of unity group-scheme over $\mathbb{Z}/p^2\mathbb{Z}$, that is $\mu_p=\operatorname{Spec}((\mathbb{Z}/p^2\mathbb{Z})[X]/(X^p-1))$. We then define the two group morphisms $$\phi_1,\phi_2:\mathbb{Z}/p\mathbb{Z}\rightarrow \mu_p(\mathbb{Z}/p^2\mathbb{Z})$$ by $\phi_1(a)=\phi_2(b)=1$. They are the zero maps. The section $1$ is the evaluation at $X=1$ ring map. We denote by $\phi$ the induced morphism $\mathbb{Z}/p\mathbb{Z}\oplus \mathbb{Z}/p\mathbb{Z} \rightarrow \mu_p(R)\oplus \mu_p(R)$ defined by $\phi(a,b)=(1,1)$ which is again the zero map.  
These group morphisms define sets of sections of $\mu_p$ and $\mu_p\times \mu_p$ simply by looking at their images. It happens that $\phi_1$ and $\phi_2$ do not define full sets of sections of $\mu _p$. Indeed, if we look at the element $f=X^p-a\in (\mathbb{Z}/p^2\mathbb{Z})[X]$, we see that the matrix of the multiplication-by-$f$ endomorphism in the free $\mathbb{Z}/p^2\mathbb{Z}$-module $(\mathbb{Z}/p^2\mathbb{Z})[X]/(X^p-1)$ of rank $p$,with basis $1,X,\ldots, X^{p-1}$, has determinant $1-a^p$, which is different from $f(1)^p=(1-a)^p$ for at least one value of $a$ in $\mathbb{Z}/p^2\mathbb{Z}$. 
In other words, we do not have the following equality of Cartier divisors in $G_m$ over $\mathbb{Z}/p^2\mathbb{Z}$: $$(X-1)^p\not = (X^p-1) \quad\text{in}\quad (\mathbb{Z}/p^2\mathbb{Z})[X]$$
However, $\phi$ actually defines a full set of sections of $\mu_p \times \mu_p$ over $\mathbb{Z}/p^2\mathbb{Z}$. To show this, we need to verify the following statement:  
For any algebra $R$ and for any $f(X,Y)\in R[X,Y]/(X^p-1,Y^p-1)$, we have the identity $\operatorname{Norm}(f)=f(1,1)^{p^2} \operatorname{mod} p^2R$.  
For this, we may view the norm as the composite of $$R[X,Y]/(X^p-1,Y^p-1)\xrightarrow{N_X}R[Y]/(Y^p-1)\xrightarrow{N_Y}R$$
First, note that in $G_m$ over $\mathbb{Z}/p\mathbb{Z}$, we do have the equality of Cartier divisors $$(X-1)^p = (X^p-1) \quad\text{in}\quad (\mathbb{Z}/p\mathbb{Z})[X]$$
Hence, the zero morphism actually defines a full set of sections of $\mu_p$ over $\mathbb{Z}/p\mathbb{Z}$. It follows that we have the norm identity for the $\mathbb{Z}/p\mathbb{Z}$-algebra $(R/pR)[Y]/(Y^p-1)$, which results in the congruence $$N_X(f(X,Y))=f(1,Y)^p+pg(Y)$$ with some $g(Y)\in R[Y]/(Y^p-1)$. 
Now, for any two elements $A,B \in R[Y]/(Y^p-1)$ and an indeterminate $T$, the norm with respect to $Y$ of $A+TB$, that is the image of $A+TB$ by the map $R[T][Y]/(Y^p-1)\xrightarrow{N_Y} R[T]$, is a polynomial in $T$ of degree $p$, of the form $$N_Y(A+TB)=N_Y(A)+\sum_{i=1}^{p-1}T^i\Lambda_i(A,B) + T^pN_Y(B)$$
On the other hand, $\operatorname{mod} p$, we have $$N_Y(A+TB)=(A(1)+TB(1))^p=A(1)^p+T^pB(1)^p \quad\operatorname{mod}p$$
Equating coefficients, we obtain $\Lambda_i(A,B)=0\quad\operatorname{mod}p$ for $i=1,\ldots ,p-1$ and therefore $N_Y(A+TB)=N_Y(A)\quad\operatorname{mod}(pT,T^p)$.  
Applying this with $T=p$, we conclude $N_Y(A+pB)=N_Y(A)\quad\operatorname{mod}p^2$.
Eventually, $$\begin{align}
\operatorname{Norm}(f(X,Y))&=N_y(N_X(f))\\&=N_Y(f(1,Y)^p+pg(Y))\\&=N_y(f(1,Y)^p)\quad&\operatorname{mod}p^2\\&=N_Y(f(1,Y))^p \quad&\operatorname{mod}p^2\\&= (f(1,1)^p+pk)^p \quad&\operatorname{mod}p^2\\&= f(1,1)^{p^2}\quad&\operatorname{mod}p^2 
\end{align}$$
