1

I need to find all the common roots of the two polynimials $10x^{15}+9x^2+1$ and $10x^{15}+8x^2+2$ over $GF(19)$. 1 is not a root of either. The only method that I can come up with is trying brute force to just divide the polynomials by $x-r$ where $r \in GF(19)$. What am I not seeing here that I can only come up with this method? What did I not understand?

In general, is there an efficient way to find the roots of a polynomial over a finite field?

Thank you for your help.

DariusTheGreat
  • 426

1 Answers1

3

Certainly $x=-1$ is a common root, since $f(-1)=g(-1)=0$ in $\mathbb{F}_{19}$. Suppose that $a$ is another common root. Then $x-a$ divides $f(x), g(x)$ and hence also $$f(x)-g(x)=x^2-1=(x-1)(x+1).$$ Since $1$ is not a common root, we are done. With more work we also see that $\gcd(f(x),g(x))=x+1$. Here $f(x)$ has another linear factor $x+3$, and $g(x)$ another linear factor $x+10$.

Edit: $f(x)=10x^{15}+9x^2+1$ and $g(x)=10x^{15}+8x^2+2$.

user26857
  • 52,094
Dietrich Burde
  • 130,978
  • if you compute f(1) you get 10 + 9 + 1 = 20 which is 1 in $GF(19)$ not zero.. – DariusTheGreat Jul 06 '17 at 15:02
  • 1
    But I meant $f(-1)=-10+9+1=0$. – Dietrich Burde Jul 06 '17 at 15:03
  • No more work s needed, since $,x+1\mid d\mid x^2-1$ and $,x-1\nmid d,\Rightarrow, d = x+1.\ \ $ – Bill Dubuque Jul 06 '17 at 15:12
  • I get it now thanks. There's a Lemma in my textbook which states: in any commutative ring, if a divides b and a divides c then a also divides (b+c). That's the lemma we use here. Why do we get (x-1) as a factor although it's not a common root? since (x-1) divides f(x)-g(x) shouldn't 1 also be a root? – DariusTheGreat Jul 06 '17 at 15:12
  • 1
    No, if $p(x)$ divides $f(x)-g(x)$, then it does not follow in general that $p(x)\mid f(x)$, or $p(x)\mid g(x)$. Image integers: $2\mid (5-3)$, but both $5$ and $3$ are not divisible by $2$. – Dietrich Burde Jul 06 '17 at 15:15