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

Question

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.

Please explain me.

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@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

OhMyGod · Accepted Answer · 2013-04-30T15:35:01.123

14

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.

edited Apr 30 '13 at 15:35

answered Apr 30 '13 at 15:24

OhMyGod

509

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
2

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
1

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
@NewDeveloper if it is commutative, then only one scenario is enough. – OhMyGod Apr 30 '13 at 15:34
@DonAntonio i edited that. Thanks and thanks! :) – OhMyGod Apr 30 '13 at 15:35
That should be enough to make this response a better answer. ;) – NasuSama Apr 30 '13 at 15:36
Thanks @OhMyGod. Also the buddies who gave their options :) – New Developer Apr 30 '13 at 15:39
Wow... I got a silver badge before realizing what badges are '-' – OhMyGod Apr 30 '13 at 17:11

score 3 · Answer 2 · answered Apr 30 '13 at 15:26

3

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).

answered Apr 30 '13 at 15:26

Hagen von Eitzen

374,180

score 0 · Answer 3 · answered Nov 01 '23 at 22:48

Similar to the answer of Hagen von Eitzen above (I credit his work), it is a well-known result if $F$ is a field, then $F[x]$ is Euclidean domain. And (given a irreducible polynomial (irreducible in F[x]), say $r(x)$ in $F[x]$) if $D=F[x]/<r(x)>$, then $D$ is a field. As a example, $R[x]/<x^2+1>$ is field (can you see which field)? Now, for proving the final result, all that we need to prove is (over $F_2$), $x^2+x+1$ is irreducible. As $x^2+x+1$ is a degree $2$ polynomial, we merely need to show $x^2+x+1$ has (in $F_2$) no roots. Thus...

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

3 Answers3

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