# Polynomial factorization to irreducible factors with respect to field

I have a question, I think I don't understand this material very well and could use an explanation / some help.

Basically we are asked to decompose $x^5-x$ to irreducible factors over $R,F2,F5,C$

But I don't understand how the field has anything to do with it?

I would just say: $x^5-x = x(x^4-1) = x(x^2-1)(x^2+1) = x(x-1)(x+1)(x^2+1)$ and that's it, I can't factor it anymore. I don't see how fields are related.

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Over $\mathbb{C}$, $x^2+1$ factors as $(x+i)(x-i)$. Over $F_2$, it factors as $(x+1)(x+1)$. Over $F_5$, it factors as $(x+2)(x+3)$.

Remark: Polynomials of degree $1$ are irreducible, so it's over. Note that any polynomial with complex coefficients of degree $\ge 1$ can be expressed as a product of polynomials of degree $1$ and coefficients in $\mathbb{C}$. This is a version of the Fundamental Theorem of Algebra.

For $F_2$, note that $1^2+1=0$. So by a generalization of a familiar result, $x-1$ divides $x^2+1$ over $F_2$. We gave the polynomial $x-1$ the equivalent name $x+1$.

For $F_5$, note that $(2)^2+1=0$. So over $F_5$, the polynomial $x-2$, that is, $x+3$, divides $x^2+1$.

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Is there any algorithm / way to determine how a polynomial factors in a field? or did you just use common sense and experience. because I'm new at this, I won't intuitively think that over F5 (x+2)(x+3) is equal to $x^2+1$. But great answer thanks! – Oria Gruber Nov 13 '13 at 21:34
I added some information about to get the result for $F_5$. There are algorithms. Since the subject is important, lots of work has been done. For polynomials of degree $2$ or $3$, it is easy, such a polynomial is irreducible iff it has no root in the field. Higher degree polynomials are much messier. – André Nicolas Nov 13 '13 at 21:39
ok, next question, i am asked to factor $x^2+x+1$ in $R,F2$. I want to say that in $R$ I can't factor it since no real number is a root of that polynomial. is that correct? and don't tell me the answer to $F2$ I will figure that out myself – Oria Gruber Nov 13 '13 at 21:45
Yes, your reasoning is right for the reals. As to the answer that you don't want, I have partly given it in the comment just above yours. – André Nicolas Nov 13 '13 at 21:48
it's also impossible to factor it in $F2$ since it does not have any roots in $F2$. – Oria Gruber Nov 13 '13 at 21:52