Question: > 
> Show that if the encryption exponent $3$ is used for the RSA cryptosystem by three different people with different moduli, a plaintext message $P$ encrypted using each of their keys can be recovered from these resulting three ciphertext messages.
> 
> 
> 

*Proof*. We have
$$x\_1\equiv P^3\pmod{n\_1},$$
$$x\_2\equiv P^3\pmod{n\_2},$$
$$x\_3\equiv P^3\pmod{n\_3},$$
and using the Chinese Remainder Theorem, we can get
$$P^3\equiv x\_1(n\_2n\_3)(n\_2n\_3)^{-1}+x\_2(n\_1n\_3)(n\_1n\_3)^{-1}+x\_3(n\_1n\_2)(n\_1n\_2)^{-1}\pmod{n\_1n\_2n\_3}.$$
Therefore, $P$ can be found by testing all cubes that satisfy this congruence. $\square$

Does this proof seem rigorous enough, or should I step it up one more notch and compute for $P$? Or is perhaps the exercise asking for something else? What do you guys think? Thanks in advance!

Answer: 
What is your definition of trace? Best one is the evaluation map:
$$
\phi:M\rightarrow M \text{ is an element of } M\otimes M^\vee
$$
because for $M$ free
$$
M\otimes M^\vee \rightarrow \operatorname{End(M)} \quad m\otimes \phi \mapsto \phi(-)m
$$
is an isomorphism. Trace is the evaluation
$$
\operatorname{tr}: M\otimes M^\vee \rightarrow R
$$
Hint:

Let $\{a\_i\}$ be a basis for $M$ and $\{b\_j\}$ a basis for $N$. Prove then that $[f\otimes g]\_{\{a\_i\otimes b\_j\}}=[f]\_{\{a\_i\}}\otimes [g]\_{\{b\_j\}}$ where by $[\cdot]\_\mathscr{B}$ I mean matrix representation with respect to $\mathscr{B}$ and the tensor product on the right hand side is the [Kronecker product](http://en.wikipedia.org/wiki/Kronecker_product).