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I just need a check for my reasoning:

If I have a ring $R=\mathbb{F}_p[X]$ and elements $x,y\in R$ it is true $$(x+y)^p=x^p+y^p?$$ It is true since every other element of expansion has a multiple of $p$ as coefficient.

What about field $R=\mathbb{F}_p[X]/(f)$ where $f$ is polynomial of degree $n$. I think rule still holds for the same reason.

Also what can be said about $x^p$? Is there any general rule about that element? It seems to me that $x^p=x$ since $(\mathbb{F}_p,\cdot)$ has order $p-1$. Am I missing something?

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    Very bad idea to use $\;p\;$ for two very different things...There are plenty of unused letters out there. – 2017-01-24
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    In any commutative ring where p=0 you have that $(a+b)^p=a^p+b^p$ – 2017-01-24
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    Yes $R = \mathbb{F}_p[X]/(f)$ is a ring of characteristic $p$, so that $(a+b)^p = a^p + b^p$ for any $a,b \in R$. If $f \in \mathbb{F}_p[X]$ is irreducible then $R$ is a finite integral domain and hence a field, with $q= p^{deg(f)}$ elements. Otherwise factorize $f = \prod_{m=1}^n f_m$ and $R \simeq \mathbb{F}_{q_1}\times \ldots \times\mathbb{F}_{q_n}$ where $q_m = p^{deg(f_m)}$. So $R$ is a direct product of some fields of characteristic $p$, thus it is a non-integral domain of characteristic $p$. – 2017-01-24
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    @DonAntonio: I didn't realize it at first. Changed polynomial to f instead of p – 2017-01-24

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As has been noted in the comments, for any commutative ring of characteristic $p$ where $p$ is prime, you have for all $x,y$, $(x+y)^p = x^p + y^p$. Where I come from it's called "the undergrad's dream". However, $x^p = x$ does not hold for any $x$. In fact, if $R$ is a commutative integral domain with subring $\mathbb{F}_p$, the only $x$'s such that $x^p = x$ are elements of $\mathbb{F}_p$ (that's because you have euclidian division by $X-k$ in $R[X]$, and so an easy argument on polynomial roots works here) I don't know whether the hypotheses I've given are minimal, but they're enough for what you're asking.

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    Great answer :) I think that's it. But what about special ring for polynomial $f=x^2$. Then elements of $R$ are of the form $ax+b$ and $(ax+b)^p=a^px^p+b^p=ax+b$ since $p$ is odd and $x^2=0$? – 2017-01-24
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    Well first of all $R$ is not an integral domain. Moreover, $p$ being odd and $x^2 =0$ gives $(ax+b)^p = b^p$ as $x^p = x^2 * x^k = 0$. Now $b\in \mathbb{F}_p$ gives $b^p = b$ so $(ax+b)^p = b$. – 2017-01-24