Jump to content

Recommended Posts

Posted

I have seen asserted that for polynomials f and h, f divides h implies that fbar divides hbar, where "bar" indicates reduction modulo p, a prime number say.

 

But when I test this with sample polynomials, it does not seem to be true. For example modulo 3 and f = 2x + 1, h = 6x2 + 7x + 2, then fbar = 2x + 1, hbar = x + 2.

What is wrong here please ?

 

Posted

Thank you. But how about this case (mod 3) :

f = 11x + 2, g = 5x + 7, then fg = 55x2 + 87x + 14, fbar = 2x + 2, gbar = 2x + 1, (fg)bar = x2 + 2.

How can fbar (or gbar) divide (fg)bar ?

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.