In the above picture of example 11.6.3, my questions are:
1.) why is both the order of $eR’$ and $e’R’$ $11$? Does it have something to do with $11$ being prime and that $122$ = $11 \times 11$?
2.) How do we know $eR’$ and $e’R’$ are isomorphic to $F_{11}$
Thanks!
Looking at the text, it seems the author is using the following very elementary argument.
We have two idempotents $e,e'$ such that $e+e'=1$, and hence also that $ee'=0=e'e$ (seen for example by multiplying $e+e'=1$ on the left or right by $e$). Since the ring $R'$ is commutative, it follows that $R'\cong R'e\times R'e'$. Now $R'e$ and $R'e'$ are both non-trivial rings, with respective units $e,e'$, and $11^2$ has only one proper factorisation, so necessarily $|R'e|=11=|R'e'|$. Finally, since 11 is prime, $R'e$ is generated as an additive group by the unit $e$. Thus the ring homomorphism $\mathbb Z\to R'e$ is onto, and yields an isomorphism $\mathbb F_{11}=\mathbb Z/(11)\xrightarrow\sim R'e$.
The more advanced way of seeing this goes by way of the Chinese Remainder Theorem, which in one form says that if we have a (commutative) ring $R$ and ideals $I,J$ such that $I+J=R$, then $I\cap J=IJ$ and there is an isomorphism $R/IJ\xrightarrow\sim R/I\times R/J$.
For your question, we would take $R'=\mathbb F_{11}[t]/(t^2-3)$, which is what we mean when we say we abstractly/artificially adjoin a square root of 3. In $\mathbb F_{11}[t]$ we have the ideals $I=(t-5)$ and $J=(t+5)$, so that $I+J=\mathbb F_{11}[t]$, and also $\mathbb F_{11}[t]/I\cong\mathbb F_{11}$ via the map $t\mapsto 5$. Similarly for $J$, and so the Chinese Remainder Theorem gives $$ R'\cong\mathbb F_{11}[t]/IJ\cong\mathbb F_{11}[t]/I\times\mathbb F_{11}[t]/J\cong \mathbb F_{11}\times\mathbb F_{11}. $$ It is easy to check that $e$ is sent to $(0,1)$ and $e'$ is sent to $(1,0)$.