My question refers to a step in a proof from J. Milne's "Etale Cohomology" treating equivalent statement about Henselian rings:
In the proof (a) implies (b) he firstly shows that if $B$ has the shape $B=A[T]/(f)$ then $B \cong \prod B_{m_i}$ for $m_i$ maximal primes of $B$.
Then the author considers the case where $B$ is not local (so by remark $B/mB$ and going up also not local). Then we choose a $b \in B$ such that $\overline{b} \in B/mB$ is nontrivial idempotent in $B/mB$
First question: can we always find such $$\overline{b}$ in non local rings?
Futhermore by mapping $T \mapsto b$ we can declare a morphism $\phi: A[T]/(f) =:C \to B$.
My second QUESTION is why we can always find an idempotent $c \in C$ with $\bar{\phi(c)}=\bar{b}$?
So why is it idemponent, so $c^2=c$?
Here are three facts one encounters in an introductory graduate algebra class which make answering this question straightforwards:
Fact 1. Every artinian ring is a finite product of local artinian rings.
Fact 2. A product decomposition of a ring $R=R_1\times R_2\times \cdots \times R_n$ exactly corresponds to a list of nontrivial idempotents summing to $1$, and the idempotents are exactly the $1$ in each $R_i$.
Fact 3. The nonzero homomorphic image of any local ring is again a local ring.
To answer your first question, we first note that $B/mB$ is a finite module over a field, and thus is a finite-dimensional vector space. Thus it is artinian and we may apply fact 1 to see it is a product of artinian local rings. If $B/mB$ is not local, we may write it as a product of local rings, and apply fact 2 to find the nontrivial idempotents you seek.
For the second question, note that the image of $C$ contains the two distinct nontrivial idempotents $\overline{b}=\phi(T)$ and $1-\overline{b}=\phi(1-T)$. Thus the image $\phi(A[T]/(f))$ has a nontrivial product decomposition by fact 2. But this contradicts fact 3 (no local ring is a nontrivial product of rings) so $A[T]/(f)$ can't be local. By the logic of the paragraph above the red squiggle, $f=gh$ with $g,h$ strictly coprime and $A[T]/(f)\cong A[T]/(g)\times A[T]/(h)$. By fact 2, there are two idempotents in this ring: $c_1,c_2$ so that $c_1=0\mod g, c_1=1\mod h$ and $c_2=1\mod g,c_2=0\mod h$. Depending on whether $g(b)=0$ or $h(b)=0$ (remember, only one of these can be true by strict coprimality), which one of $c_1,c_2$ to take as $c$ changes - I'll leave this as an easy exercise for the reader.
Finally, this screenshot you've posted seems to be from an older version of Milne's book - the latest version is on his website and may be put together in such a way to cause you fewer questions.