Proof of the Fundamental Theorem of Algebra for Congruent Equations 📂Number Theory

Proof of the Fundamental Theorem of Algebra for Congruent Equations

Theorem ¹

For a given prime number $p$, if we denote $p\nmid a_{ 0 }$, polynomials $$ f(x)=a_{ 0 }x^{ d }+a_{ 1 }x^{ d-1 }+ \cdots +a_{ d-1 }x+a_{ d } $$ with all integer coefficients have at most $d$ distinct solutions to the equation $f(x)\equiv 0 \pmod{p}$.

Explanation

Just as commonly known, for polynomials with real coefficients, a $n$-degree equation includes $n$ solutions, including multiplicity. From the perspective of number theory, this implies that a $d$-degree equation with only integer coefficients has at most $d$ solutions. This statement can be further simplified by introducing complex numbers to integers.

Proof

Strategy: Contrary to how the Fundamental Theorem of Algebra is derived through complex analysis, elementary number theory knowledge is sufficient for congruence equations. Assume that there are more than $d$ distinct solutions to induce a contradiction.

Assume there exists an $n$-degree integer coefficient polynomial $P$ that has more distinct solutions than $n$ as given by equation $P(x)\equiv 0 \pmod{p}$. Among them, choose the polynomial $f$ with the smallest degree, which can be represented as follows: $$ f(x)=A_{ 0 }x^{ d }+A_{ 1 }x^{ d-1 }+ \cdots +A_{ d-1 }x+A_{ d } (p\nmid A_{ 0 }) $$ Then, equation $f(x)\equiv 0 \pmod{p}$ has distinct solutions $r_{ 1 },r_{ 2 }, \cdots ,r_{ d },r_{ d+1 }$. For the solution $r$ of $f(x)\equiv 0 \pmod{p}$, $$ f(x)=(x-r)g(x)+f(r) $$ is represented as above. The quotient $g(x)$ of $f(x)$ divided by $(x-r)$ is a $(d-1)$-degree integer coefficient polynomial as follows: $$ g(x)=B_{ 0 }x^{ d-1 }+B_{ 1 }x^{ d-2 }+ \cdots +B_{ d-2 }x+B_{ d-1 } (p\nmid B_{ 0 }) $$ Substituting $r$ into $r_{ 1 }$ results in $$ f(x)\equiv (x-r_{ 1 })g(x) \pmod{p} $$ Since $r_{ k }$ is a root of $f(x)=0$, for $$ f(r_{ k })\equiv (r_{ k }-r_{ 1 })g(r_{ k })\equiv 0 \pmod{p} $$ $r_{ k }-r_{ 1 } \not\equiv 0 \pmod{p}$ implies $g(r_{ k })\equiv 0 \pmod{p}$ must be true. Thus, $g(x)\equiv 0 \pmod{p}$ has $d$ distinct solutions $r_{ 2 },r_{ 3 }, \cdots ,r_{ d+1 }$. Since $f(x)$ is the lowest degree polynomial among those with more distinct solutions than its degree, namely $d$-degree, and $g(x)$, despite being $(d-1)$-degree, has $d$ distinct solutions, it leads to a contradiction.

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