Isomorphism

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November 13th 2005, 03:29 PM
Treadstone 71

Isomorphism

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

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.
November 13th 2005, 08:34 PM
hpe

Quote:

Originally Posted by Treadstone 71

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

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 $\mathbb{Z}_5 \times \mathbb{Z}_5$ , so you can use the bijection $\Phi: ax+b \mapsto (a,b)$ to lift the multiplication on $\mathbb{Z}_5[x]/(x^2+1)$ to $\mathbb{Z}_5 \times \mathbb{Z}_5$ . For a different quadratic polynomial, you would get a different ring structure (a field for $\mathbb{Z}_5[x]/(x^2+2)$ ).
November 14th 2005, 03:43 AM
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)?
November 14th 2005, 02:51 PM
hpe

Quote:

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 $\Phi: Z_5[x]/(x^2+1) \to Z_5 \times Z_5$ by $\Phi(ax+b) = (a,b)$ , where $ax+b$ is the unique representative of degree less than 2 of an element in $Z_5[x]/(x^2+1)$ . This is a bijection. Then define
$(a,b) \cdot (c,d) = \Phi\left( \Phi^{-1}(a,b) \cdot \Phi^{-1}(c,d) \right)$ , where the multiplication on the right is the one in $Z_5[x]/(x^2+1)$ . Then work out what this means in $Z_5 \times Z_5$ . Compare this to the formula for complex multiplication, and think about an explanation.

Hope this helps.