# congruence classes and roots of a polynomial.

• Mar 4th 2011, 07:35 AM
hmmmm
congruence classes and roots of a polynomial.
I am reading a text book on elementary number theory and I am reading a chapter on congruence classes at the moment and im confused as to what it means when it asks/says that a polynomial \$\displaystyle f(x)=x^2+1 \$ has two roots in Z_2 but none in Z_3?

Thanks very much for any help.
• Mar 4th 2011, 07:45 AM
kira
edited
• Mar 4th 2011, 07:47 AM
TheEmptySet
Quote:

Originally Posted by hmmmm
I am reading a text book on elementary number theory and I am reading a chapter on congruence classes at the moment and im confused as to what it means when it asks/says that a polynomial \$\displaystyle f(x)=x^2+1 \$ has two roots in Z_2 but none in Z_3?

Thanks very much for any help.

A root is a value of x that make the polynomial equal to zero.

In \$\displaystyle \mathbb{Z}_{2}\$ there are only two equivalence classes.

\$\displaystyle [0],[1]\$

\$\displaystyle f([0])=[0]^2+[1]=[1] \ne [0]\$ and

\$\displaystyle f([1])=[1]^2+[1]=[1] +[1]=[2] =[0]\$ so it is a root and the polynomial can be factored mod 2 as

\$\displaystyle (x+[1])(x+[1])=x^2+x+x+1 =x^2+[2]x+1=x^2+1\$

Now try the same thing in \$\displaystyle \mathbb{Z}_3\$
• Mar 4th 2011, 07:53 AM
hmmmm
Thanks very much I understand now i was just getting a bit confused.
• Mar 5th 2011, 06:19 AM
HallsofIvy
For those of us who prefer very strict language, a polynomial does NOT have "roots", equations have roots. Better terminology is that the zeros of a polynomial, p(x), are the roots of the equation p(x)= 0.
• Mar 7th 2011, 02:16 AM
hmmmm
Thanks, I probably do need to be a lot more precise with my notation/terminology