# Seeing that $\Bbb F_2[x]/(x^2+x+1)$ is a field

I have some basic question with polynomials appreciate if someone could explain me this.

Following is additional and multiplication tables and it is say that this is a field. Have no idea why say it is field.

-
Do you know what a field is? – Rhys Apr 30 '13 at 15:23
@Rhys- I know very less. The definitions are very difficult to understand me. – New Developer Apr 30 '13 at 15:25
See the answers below to understand what a field is. Even if the notations are a bit complicated for you to understand, you should get the hang of what the answerers explain to you. – NasuSama Apr 30 '13 at 15:27

Just note that, for multiplication, every non-zero element on the first column has, on its line, a $1$, which means it has a inverse to the right... Now, for every element on the first row, which contains nonzero elements, it has a 1 on its column, which means it is left-invertible.

Thus, it is a field since every nonzero element is invertible.

-
You made a good point! – NasuSama Apr 30 '13 at 15:25
And I just entered Math Stack Exchange! :P Begginer's luck I guess. – OhMyGod Apr 30 '13 at 15:25
Don't worry about this. Everyone has the first time. Nobody can go far beyond the "StackExchange" habitants, or people who go on that site for very long time (Know: experience!) Good job posting the great answer! – NasuSama Apr 30 '13 at 15:26
Last line must be: "...since every nonzero element is invertible"...just nitpicking, and welcome! +1 – DonAntonio Apr 30 '13 at 15:32
In order to become a field, the two scenarios (row wise and column wise) should be there or having 1 is enough? – New Developer Apr 30 '13 at 15:33

If $R$ is a ring with unity and $f\in R[x]$ a polynomial, then $R[x]/\langle f\rangle$ is always at least a ring. To be a field, it needs a unity (check) and each nonzero element must have a multiplicative inverse (check).

-