Polynomial factorization to irreducible factors with respect to field

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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 $x^2+1$.