Polynom equality modulo p

Question

I found these two equations:

(a) $$X^4 + 1 \equiv (X + 1)^4 \mod \ 2$$ (b) $$X^4 + 1 \equiv (X^2 - X - 1)(X^2 + X - 1) \mod \ 3$$

I would like to understand the concept of modulo for Polynoms.

How were they made? And how to verify these equality ?

Thanks in Advanced!

Answer 1

Note that $\;2k=0\pmod 2\;,\;\;\forall\,k\in\Bbb Z\;$ , and thus

$$(x+1)^4=x^4+\overbrace{4x^3+6x^2+4x}^{=0\pmod 2}+1=x^4+1\pmod 2$$

Try now something similar with the other case

Answer 2

It seems your question is: how does integer congruence arithmetic ("modular" arithmetic) extend to polynomials? Simply because the proofs of the basic congruence laws (sum and product rules) work universally (technically: they use only basic ring laws (distributive,commutative,associative), so the proofs remain valid in any commutative ring). This universality becomes clearer when one studies abstract algebra, which studies more structural versions of these ideas (ideals & quotient rings, a.k.a. residue rings).

In particular $\ f\equiv g\pmod m\ $ means that $\,m\mid f-g\,$ i.e. $\,(f-g)/m\in \Bbb Z[x],\,$ i.e. it is a polynomial with integer coefficients. Let's look at the proof of the product rule

$\bf{Congruence\ Product\ Rule}\rm\ \ \ \ \ \color{#c00}{F\equiv f},\ \ and \ \ \color{#0a0}{G\equiv g}\ \Rightarrow\ \color{blue}{FG\equiv fg}\ \ \ (mod\ m)$

$\bf{Proof}\rm\ \ \ m\mid \color{#c00}{F\!-\!f},\ \color{#0a0}{G\!-\!g}\ \Rightarrow\ m\mid (\color{#c00}{F\!-\!f})\ G + f\ (\color{#0a0}{G\!-\!g})\ =\ \color{blue}{FG - fg}\quad $ QED

This is the same as the proof for integers, except that we are working with polynomials. The proof works in both cases because it uses only basic (ring) arithmetic laws (distributive,commutative, associative), which hold true both for integers $\,\Bbb Z\,$ and polynomials $\,\Bbb Z[x],\,$ since both are commutative rings.

In particular, when computing mod $m$ the Product Rule yields $\,i\equiv j\,\Rightarrow\, i x^k\equiv j x^k.\,$ So we can replace all coefficients by their remainder mod $m$, i.e. we can mod out coefficients. Doing that in both of your examples quickly yields the sought congruences.

Remark $\ $ Informally (in rings) working "mod m" means adjoining the hypothesis $\,m = 0$ to the ambient (ring) axioms. Applying the ring axioms then yields inferences that other elements $= 0,\,$ e.g. if $\,m = 0 = m'\,$ that $\,am+b m' = 0\,$ for all $\,a,b\,$ in the ring. The algebraic structure underlying these matters will become clearer when one studies ideals of rings.

Answer 3

It just results from the fact that in a (commutative) ring of characteristic $2$ (i.e. a ring $R$ for which $2\cdot 1_R=0_R$) the map $x \mapsto x^2$ is a ring homomorphism (called the Frobenius morphism): for all $x,y\in R$ $$(xy)^2=x^2y^2,\quad (x+y)^2=x^2+y^².$$ The last equality results from the binomial formula, noting that $2xy=0$ since the characteristic is $2$. Now an equality between polynomials modulo $2$ means we're looking at polynomials with coefficients in the field $\mathbf F_2=\mathbf Z/2\mathbf Z$, which is of characteristic $2$.

By a trivial induction, we also have $$(x+y)^{2^n}=x^{2^n}+ y^{2^n}.$$

Answer 4

For $(a),$

$y^{m+1}-y=y(y^m-1)$ is divisible by $y(y-1)$ which being the product of two consecutive integers is even

$\implies y^{m+1}\equiv y\pmod2$ for $m\ge1$

$\implies y^4\equiv y\pmod2$ set $y=X, X+1$

For $(b),$

$(x^2-1-x)(x^2-1-x)=(x^2-1)^2-x^2=x^4-3x^2+1\equiv x^4+1\pmod3$