
Polynomial Rings
How is this true?
6x^3 + 2x^2 x + 1 = x^2 +2x + 1 in F (sub3) [x]
There is a dash over the 6, 2, 1 in the first polynomial and a dash over the 2, 1 in the second
I'm thinking original algebra days and obviously thats not what were doing here. Can anyone explain how to figure the above.

$\displaystyle F_3$ is the finite field of order 3. It is just the integers modulo 3.
$\displaystyle a \equiv b $ (mod n) iff n(ab)
$\displaystyle F_3[x]$ is simply polynomials with coefficients coming from $\displaystyle F_3$.
Thus in $\displaystyle F_3$
$\displaystyle 6 \equiv 0$ mod 3
as 360=6
and
$\displaystyle 2 \equiv 1$ mod 3
32(1)=3

Okay...that makes sense, but where does the 2x (with a bar over the 2) come from?
Can you go through each step on how to come up with:
x^2 +2x + 1 (bars over the 2 and 1)
Do I only look at the numbers with a bar over them when using the mod formula?
Thanks...

The bars are talking about the equivalence classses of the integers modulo 3. There should be a bar over every coefficient in the polynomial ring. It is just because the leading coefficient was written $\displaystyle x^2$ instead of $\displaystyle \bar{1}x^2$. And since $\displaystyle \bar{6} = \bar{0}$ they just left off that term.

Awesome...You;re great!
One more thing...The final term of 6x^3 + 2x^2 x + 1 is obviously 1.
How come the 1 doesn't change?

well certainly 1 is congruent to 1 mod 3.
Like don't think this polynomial on the right is some special form of the polynomial or anything. It is not like every polynomial should be reduced into this particular form. If anything you should always reduce it to the form where you have the coefficients represented by the least positive residue.
That is a polynomial over the integers mod p, represented by 0,1,2,..., p. One of those.
This is just asking you to notice that these two things are the same because all the coefficients are congruent mod p.