# Polynomials over GF(7)

The following exercise is from Golan's book on linear algebra.

Problem: Consider the algebra of polynomials over $GF(7)$, the field with 7 elements.

a) Find a nonzero polynomial such that the corresponding polynomial function is identically equal to zero.

b) Is the polynomial $6x^4+3x^3+6x^2+2x+5$ irreducible?

Work so far: The first part is easy. The polynomial $x^7-x$ works by Fermat's little theorem. The second part is trickier. If the polynomial is reducible, it facts into the product of a linear term and something else, or it factors as two quadratics. The first case is easy to exclude; simply plug all seven elements of $\mathbb{Z}_7$ into the polynomial and confirm none of them are a root. The second is harder. Of course, one could just set up the systems of equations resulting from

$$(ax^2+bx+c)(dx^2+ex+f)$$

and go through all the possible values of $a,c,d,f$ and see if the resulting values of $b$ and $e$ are permissible, and while I know that would eventually give me the answer, I have no desire to do all of those computations. Is there a slicker way?

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Guess-and-check is really the best way to go here. But instead of picking values for the coefficients, I prefer to just find all the irreducible quadratics (checking, by exhaustion, to see which ones have roots) and then see if any of them divide the given polynomial. – Brett Frankel May 31 '12 at 2:59
Rather than find all $21$ irreducible quadratics, it should be easier to consider $x^{49}-x$ (the product of all the monic irreducible polynomials of degree $1$ and $2$) and use Euclid's algorithm to take the GCD of it with your polynomial. – Chris Eagle May 31 '12 at 3:03
Switch the signs, we are trying to see whether $x^4+4x^3+x^2+5x+2=(x^2+bx+c)(x^2+ex+f)$. Not a lot of work. – André Nicolas May 31 '12 at 3:31
@Potato: Then you can multiply the first by $4$ and the second by $2$, and get a decomposition in which the coefficient of $x^2$ in each factor is $1$. – André Nicolas May 31 '12 at 3:36
@Jyrki: Fortunately, we aren't trying to find the factorization, so that's not a problem. – Chris Eagle May 31 '12 at 10:33

Hint $\$ As suggested you could use trial-and-error, and program a computer to test the $7^4 = 2401$ cases that arise from $4$ undetermined coefficients in a factorization into two quadratics. But, as is often true, a little insight trumps brute force. By exploiting innate symmetry, we can reduce the $2401$ cases to $2$ cases. First, shifting $\rm\:x\to x\!-\!1\:$ to kill the $\rm\:x^3\:$ term yields

$$\rm\begin{eqnarray} -f(x\!-\!1) &\equiv&\rm\ \ \ x^4\ +\ 2\ x^2\ -\ 3\ x\ +\ 2\pmod 7 \\ &\equiv&\rm\ (x^2\!- a\, x + b)\ (x^2\! + a\, x + c)\\ &\equiv&\rm\ \ x^4\! + (b\!+\!c\!-\!a^2)\!\: x^2\! + a(b\!-\!c)\:\!x + bc\end{eqnarray}$$

Up to $\rm\, b,c\,$ swaps, $\rm\: bc\equiv 2\!\iff\! (b,c)\, \equiv\, \pm(2,1),\, \pm\:\!(3,3).\:$ $\rm\:b\not\equiv c\:$ else coef of $\rm\,x\,$ is $\,0\not\equiv -3$.

If $\rm\ (b,c) \equiv\ \ \: (\ 2,\ 1\ )\$ then $\rm\:-3 \equiv a(b\!-\!c)\equiv\ \: a\:\$ so $\rm\:b\!+\!c\!-\!a^2\equiv\ \ \ \, 2\!+\!1\!-\!(-3)^2\equiv\ \ 1\:\not\equiv 2$

If $\rm\ (b,c) \equiv (-2,\!-1)\:$ then $\rm\:-3 \equiv a(b\!-\!c)\equiv -a\:$ so $\rm\:b\!+\!c\!-\!a^2\equiv\, -2\!-\!1\!-\!(+3)^2\equiv -5 \equiv 2$

So $\rm\:a,b,c \equiv 3,-2,-1,\:$ is a solution, which yields the factorization $$\rm -f(x\!-\!1)\, \equiv\, x^4 + 2\,x^2 - 3\,x+2\, \equiv\, (x^2-3\,x-2)(x^2+3\,x-1)\pmod 7$$

Therefore $\rm\:f(x)\:$ is reducible since $\rm\:x\to x\!+\!1\:$ above yields a factorization of $\rm\:-f(x).\ \$ QED

Remark $\$ Alternatively, you could use the Euclidean algorithm to compute $\rm\:gcd(f(x\!+\!c),x^{24}\!-\!1)\:$ for random $\rm\:c,\:$ which, $\,$ for $\rm\:c=1\:$ quickly yields $\rm\:x^2\!+\!2\:|\:f(x\!+\!1),\:$ hence $\rm\:f(x)\:$ has the factor $\rm\:(x\!-\!1)^2\!+\!2\, =\, x^2-2\,x+3.\:$ This is how some factoring algorithms work.

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The answer for (a) is to use Fermat's Little Theorem for the coprime case, ie we know that $$x^7 \equiv x \pmod{7}$$ so $$x^7 - x \equiv 0 \pmod{7}.$$ The noncoprime case, ie $x \equiv 0 \pmod{7}$ is clear.

For the second answer, the best way really is guess and check (there are more complicated algorithms though). It factors into $$6(x^2 + 5x + 3)(x^2 + 6x + 3).$$

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A brute-force "guess and check" would require checking $2401$ cases! But one can prune the search space by exploiting innate symmetry - see my answer. – Bill Dubuque May 31 '12 at 18:57
I expect that a computer would make short work of it, although if one is allowed to do that then he could just call f.factor() in Sage. – Dylan Moreland May 31 '12 at 19:11