direct product of rings

• Mar 14th 2009, 04:59 AM
knguyen2005
direct product of rings
I am stuck on this question:

Show that http://www.mathhelpforum.com/math-he...bef9a48b-1.gif when http://www.mathhelpforum.com/math-he...7606b7b8-1.gif is any field in which http://www.mathhelpforum.com/math-he...bc11212c-1.gif.

I don't know how to prove that a field is isomorphic to direct product.
What properties do I need to use to prove it?

Can anyone kindly give me some instructions how to do this question.?

Thanks alot
• Mar 14th 2009, 07:44 AM
Opalg
Quote:

Originally Posted by knguyen2005
I am stuck on this question:

Show that http://www.mathhelpforum.com/math-he...bef9a48b-1.gif when http://www.mathhelpforum.com/math-he...7606b7b8-1.gif is any field in which http://www.mathhelpforum.com/math-he...bc11212c-1.gif.

I don't know how to prove that a field is isomorphic to direct product.
What properties do I need to use to prove it?

Can anyone kindly give me some instructions how to do this question.?

The ring $\displaystyle \mathbb{F}(x)/(x^2-1)$ consists of elements of the form a+bx (where $\displaystyle a,b\in\mathbb{F}$), with algebraic operations $\displaystyle (a+bx) + (c+dx) = (a+b) + (c+d)x$ and $\displaystyle (a+bx)(c+dx) = (ac+bd) + (ad+bc)x$. You have to think of a map from $\displaystyle \mathbb{F}(x)/(x^2-1)$ to $\displaystyle \mathbb{F}\times\mathbb{F}$ that preserves these operations.

The obvious map $\displaystyle a+bx\mapsto(a,b)$ preserves addition but not multiplication (because $\displaystyle (a,b)(c,d) = (ab,cd) \ne (ac+bd,ad+bc)$). So you need to think of a different way of mapping $\displaystyle \mathbb{F}(x)/(x^2-1)$ to $\displaystyle \mathbb{F}\times\mathbb{F}$. I can't say more than that without giving away the whole solution.