Just working on some exam prep questions, and I'm a bit stuck on this one.
Let $ \mathbb{F} = \{ a + bX + cX^2 | a,b,c \in \mathbb{F}_2 = \{0,1\} \} $ be a ring with the operations:
Addition, defined as:
$(a + bX + cX^2) + (a' + b'X + c'X^2) = (a+a')+(b+b')X + (c+c')X^2 $
and multiplication by formally multiplying out the expressions and setting $X^3 = 1+X$
Given that $ \psi : \mathbb{F} \rightarrow \mathbb{F} $ is an isomorphism:
Where $ \psi(a + bX + cX^2) = (a + bX + cX^2)^2 $
Find the formula for a ring homorphism $\tau: \mathbb{F} \rightarrow \mathbb{F}$, where ${\tau}^2=\psi$
Since:
$$ \psi (a + bX + cX^2) = (a + bX + cX^2)^2 $$
$$ = (a + bX + cX^2)(a + bX + cX^2) = a + cX + (b+c)X^2 $$ (using that $a + a = 0$ and $a^2 = a$ for $ \forall a \in \mathbb{F}$)
I know I require a $\tau$, such that
$$\tau^2 = \tau \circ \tau = a + cX + (b+c)X^2 $$
However, I am really not sure where to start beyond guessing. I know it is not simply the identity map.
Since we are working over the field $\mathbb{F}_2$, I tried $ \tau = \psi^2 $ So that $ \tau^2 = {\psi^2}^2 = a + (b+c)X + bX^2 $. No Luck.
I'm certain this is not an exercise in guessing - but how should I approach this question?
Thanks for any help in advance!
Since $X^3-X- 1$ is irreducible in $\mathbb F_2[X]$, $\mathbb F\simeq \mathbb F_2[X]/(X^3-X-1)\simeq\mathbb F_8$. The only isomorphism over $F_8$ are identity, $a\mapsto a^2$ and $a\mapsto a^4$. In your problem $\psi$ is $a\mapsto a^2$ and $\tau: a\mapsto a^4$ satiesfies $\tau^2=\psi$.