1. ## Isomorphism

I must find an isomorphism between $\displaystyle \mathbb{Z}_5[x]/(x^2+1)$ and $\displaystyle \mathbb{Z}_5 \times \mathbb{Z}_5$.

The problem is that multiplication is defined entirely differently in the two rings. Any attempt by me to define an iso between the two breaks down at multiplication.

2. Originally Posted by Treadstone 71
I must find an isomorphism between $\displaystyle \mathbb{Z}_5[x]/(x^2+1)$ and $\displaystyle \mathbb{Z}_5 \times \mathbb{Z}_5$.

The problem is that multiplication is defined entirely differently in the two rings. Any attempt by me to define an iso between the two breaks down at multiplication.
There is no canonical multiplication on $\displaystyle \mathbb{Z}_5 \times \mathbb{Z}_5$, so you can use the bijection $\displaystyle \Phi: ax+b \mapsto (a,b)$ to lift the multiplication on $\displaystyle \mathbb{Z}_5[x]/(x^2+1)$ to $\displaystyle \mathbb{Z}_5 \times \mathbb{Z}_5$. For a different quadratic polynomial, you would get a different ring structure (a field for $\displaystyle \mathbb{Z}_5[x]/(x^2+2)$).

3. I'm not entirely sure what you mean by "lift" the multiplication. Isn't multiplication on Z5XZ5 defined as (a,b)(c,d)=(ac,bd)?

4. Originally Posted by Treadstone 71
I'm not entirely sure what you mean by "lift" the multiplication. Isn't multiplication on Z5XZ5 defined as (a,b)(c,d)=(ac,bd)?
No. Just like multiplication on RxR = C isn't defined by (a,b)(c,d) = (ac, bd) but by (a,c)(b,d) = (ac-bd, ad+bc).
Define $\displaystyle \Phi: Z_5[x]/(x^2+1) \to Z_5 \times Z_5$ by $\displaystyle \Phi(ax+b) = (a,b)$, where $\displaystyle ax+b$ is the unique representative of degree less than 2 of an element in $\displaystyle Z_5[x]/(x^2+1)$. This is a bijection. Then define
$\displaystyle (a,b) \cdot (c,d) = \Phi\left( \Phi^{-1}(a,b) \cdot \Phi^{-1}(c,d) \right)$, where the multiplication on the right is the one in $\displaystyle Z_5[x]/(x^2+1)$. Then work out what this means in $\displaystyle Z_5 \times Z_5$. Compare this to the formula for complex multiplication, and think about an explanation.

Hope this helps.