Hello

I have to prove that

$\displaystyle D:=\mathbb{F}_{11^{5*61*103}} \rightarrow \mathbb{F}_{11^{5*61*103}}$

$\displaystyle y \rightarrow y^3+3y^2+3y+3 $

is bijective. But how can I do this?

I know it has something to with the fact that all 11,5,61,103 are distinct primes so I can do something with the chinese remainder theorem that goes in the direction

$\displaystyle D \cong \mathbb{F}_{{11}^5} \oplus \mathbb{F}_{{11}^{61}} \oplus \mathbb{F}_{{11}^{103}}$ but what theorem does that say? Or how does the right version go? I can't find such a statement into my algebra notes...

Could maybe please someone give me a hint?

Regards