Irreducible polynomial over mod 2?

Question

Show polynomial x^3+x+1 is irreducible over mod 2. Also construct a field with eight elements and multplication and addition of the tables.

Is the elements 0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1 and wil the x and + tables give these results?

Is factor theorem enough to prove the polynomial is irreducible? Thanks

Spot on, just the conformation i needed. i just wrote the roots in terms of x not a

the factor theorem is f(0)=0 then x is a factor f(1)=0 then x-1 is a factor thus showing the polynomial is reducible or not

Answer 1

Hi

Working mod2 means we have a field with 2 elements: {0,1}

So if a polynomial over F_2 is reducible, it will factorize over F_2, ie, it will have at least one root in F_2.

f(0) = 1, so 0 is not a root of f(x)
f(1) = 3 = 1 mod2, so 1 is not a root of f(x)

Therefore the polynomial is irreducible over F_2.

Since f(x) is irreducible over F_2 and is of degree 3, we can use it to construct a field of 2^3 elements.

Say a is a root of f(x) {this root lies in F_8, not F_2}.
Then a^3 + a + 1 = 0
I.e. a^3 = a+1

We check thet the powers of a generate the field:
(remember we're working mod2 so 2a=0, 1 = -1 etc.)

a^1 = a
a^2 = a^2
a^3 = a + 1
a^4 = a*a^3 = a(a + 1) = a^2 + a
a^5 = a*a^4 = a(a^2 + a) = a^3 + a^2 = a^2 +a+1
a^6 = a*a^5 = a(a^2 +a+1) = a^3 +a^2+a = a^2 +2a+1 = a^2 +1
a^7 = a*a^6 = a^3 + a = 2a+1 = 1

And the eighth element is 0.

Since the polynomial x^3+x+1 is not only irreducible, but it's roots generate elements of the field F_8, this polynomial is also PRIMITIVE.

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You can construct the tables yourself. what you basically do is, for example (a^2 +a+1) + (a^2 +1) = 2a^2 +a+2 = a etc...

Answer 2

interior the question, it says irreducible polynomials and no person has confirmed that their polynomial is unquestionably irreducible. even nevertheless, it may well be superb if all of us understand what field we are working with. Rationals? integers? reals? If we are working interior the sector of reals then all polynomial of degree 5 is reducible.

Answer 3

Not all reducible polynomials have roots.

But reducible polynomials of degree 2 or 3 do indeed always have roots, since if they have non-trivial factors one of the factors has to have degree 1.