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I understand the idea for numbers. For example, for the ring Z, we can get the prime factorization of it, and that is the irreducible factorization.

But I'm struggling to apply the same thing for polynomials in general and to understand the idea.

For example, we have $x^3 + x^2 + x + 1$ in $F2[x]$

So I'm looking at the ring of polynomials with coefficient in F2 = {0,1}, correct?

How can I move on from here?

I am noting that if we look at these series:

$1/(x-1) = 1+x+x^2+x^3+...+x^n$

Can I use this fact to try and re-write the polynomial? or else, how do I go about it?

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    It simply factors (over any ring) as $(x+1)(x^2+1)$. So in characteristic $2$, it is $(x+1)^3$. – 2017-02-07

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For your example, you can do this: $$ x^3 + x^2 + x + 1 = \frac{x^4-1}{x-1} = \frac{x^4+1}{x+1} = \frac{(x+1)^4}{x+1} = (x+1)^3 $$ using that mod $2$ we have $1=-1$ and $(a+b)^2=a^2+b^2$.

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    That makes sense. So generally though when you have (x+1)^3 in F2, how do you get the intuition that you've reached the irreducible factorization? – 2017-02-07
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    @Lana, $x+1$ is irreducible. All polynomials of degree $1$ are irreducible. – 2017-02-07
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    Ok, so whenever I factor *any* polynomial in any ring, I need to write it as a product of polynomials of degree 1? – 2017-02-07
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    @Lana, not, at all. But if you reach a factorization with polynomials of degree $1$, then you know it's the irreducible factorization. – 2017-02-07