Justify true or false:
$x^6 -1$ = $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$ in $\mathbb{Z}_7$
The answer is supposed to be true but I cannot find the reason for this to be true.
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Multiply it out and you'll probably find all coefficients are divisible by 7 except the constant and leading. Anything divisible by 7 is 0 in $\Bbb Z_7$ – Gregory Grant Apr 02 '20 at 05:53
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Note also it equals $(x-1)(x+1)(x-2)(x+2)(x-3)(x+3)=(x^2-1)(x^2-4)(x^2-9)$. Which equals $(x^2-1)(x^2-4)(x^2-2)$. That should be easier to multiply out. – Gregory Grant Apr 02 '20 at 05:55
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Since $6=-1$ and $5=-2$ and $4=-3$ in $\Bbb Z_7$ it equals
$(x-1)(x+1)(x-2)(x+2)(x-3)(x+3)$
which equals
$(x^2-1)(x^2-4)(x^2-9) = (x^2−1)(x^2−4)(x^2−2)$
$ = (x^4-5x^2+4)(x^2-2)$
$ = x^6-7x^4+14x^2-6=x^6-1$
Gregory Grant
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$\;\;\;$ Generalize from $\Bbb F_7=\frac{\Bbb Z}{7\Bbb Z}$ to any prime finite field $\Bbb F_p=\frac{\Bbb Z}{p\Bbb Z}=\{\mathbf{0,\dots,p-1}\}$ and let $\alpha$ be a variable over the multiplicative group $\Bbb F_p^*:=\{\mathbf{1,\dots,p-1}\}$ so by Lagrange's theorem $f(\alpha)=\mathbf 0$ where $f(x):=x^{p-1}-\mathbf 1=(x-\mathbf 1)(\mathbf 1+x+\cdots+x^{p-2})$.
$\;\;\;$ To see $f(x)=\prod_{\alpha\neq\mathbf 0}(x-\alpha)$, factorize $f(x)$ with non-constant monic polynomials $$f(x)=F(x)\cdot\prod_{i=1}^mf_i(x)$$ and the maximum number $m,m=\deg(\prod_{i=1}^mf_i(x))\leq p-1$, of linear factors $f_i(x)$.
$\;\;\;$ We know $f(\alpha)=\mathbf 0$ but $F(\alpha)\neq\mathbf 0$ or else by The Polynomial Factor Theorem we'd have $F(x)=(x-\alpha_0)\cdot H(x)$ for some $\alpha_0\in\Bbb F_p^*$ which would imply $$f(x)=H(x)\big((x-\alpha_0)\cdot\prod_{i=1}^m f_i(x)\big)$$ violating the maximality of $m$.
$\;\;\;$ Therefore $f(x)=J(x)\cdot\prod_{\alpha\neq\mathbf 0}(x-\alpha)$ and $J(x)=\mathbf 1$ as $f_{i_{\alpha}}(\alpha)=\mathbf 0$ for some $i_{\alpha}$ while $\deg(J(x))=0$ because $$\deg(f(x))=p-1=\deg\big(\prod_{\alpha\neq\mathbf 0}(x-\alpha)\big)=\deg(J(x))+\deg\big(\prod_{\alpha\neq\mathbf 0}(x-\alpha)\big)$$.
Oliver Kayende
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There's a gap at "implying", e.g. in $\Bbb Z_8$ we have $1$ and $3$ are roots of $x^2-1$ but $x^2-1 \neq (x-1)(x-3)\ \ $ – Bill Dubuque Apr 02 '20 at 17:05
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@Bill Dubuque. The remainder theorem applies here because $\Bbb F_7$ is a field. – Oliver Kayende Apr 03 '20 at 20:31
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That we are working over a field (or domain) is the crux of the matter that I sought to emphasize. Then we can apply the multi-root extension of the Factor Theorem, which fails in non-domains. – Bill Dubuque Apr 03 '20 at 20:44
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And it's not the Polynomial Remainder Theorem that fails (that's true over any ring). Rather, what fails is $,a\neq b,\ f(a)=0=f(b),\Rightarrow, (x-a)(x-b)\mid f,,$ as I explain in the linked answer. Said equivalently, $,x-a,$ is no longer prime (that's true iff the coef ring is a domain). When it is prime, by Euclid Lemma $,x-b\mid (x-a)f,\Rightarrow, x-b\mid f,\Rightarrow, (x-a)(x-b)\mid f,,$ just like for primes $,p\neq q,\ p,q\mid n,\Rightarrow, pq\mid n,$ for integers, i.e. lcm = product for coprimes. That's the crux of the matter – Bill Dubuque Apr 03 '20 at 21:03
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@Bill Dubuque. By the remainder theorem I mean that in a polynomial ring $\Bbb F[x]$ over a field $\Bbb F$, given $f(x),g(x)\in\Bbb F[x],\deg(f(x))\geq\deg(g(x))>0,$ then there exists $h(x),r(x)\in\Bbb F[x]$ such that $f(x)=g(x)\cdot h(x)+r(x)$ and $\deg(r(x))<\deg(g(x))$ ($r(x)$ is the "remainder"). This statement is not true when we replace $\Bbb F$ with an arbitrary ring. – Oliver Kayende Apr 04 '20 at 00:21
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@Bill Dubuque. The factor theorem is the immediate corollary to the remainder theorem for the case $g(x)=x-a,f(a)=0,$ which must yield $r(x)=0$. Then the same is done to the "cofactor" $h(x)$ allowing us to extend the factor theorem to all roots of $f(x)$ including multiplicity. – Oliver Kayende Apr 04 '20 at 00:54
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The problem is that the factor theorem doesn't extend to multiple factors except over domains. This is explained at length in the post I linked above. – Bill Dubuque Apr 04 '20 at 01:01
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As for the relationship between the Factor and Remainder Theorems, that is explained here. – Bill Dubuque Apr 04 '20 at 01:04
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@Bill Dubuque. Applying the remainder theorem to $f(x),g(x)$ from my answer we see immediately that $r(x)$ must have at least $6$ roots implying $r(x)=0$ because $\deg(r(x))<\deg(g(x))=6$ yielding $f(x)=\lambda\cdot g(x)$ as in my proof. – Oliver Kayende Apr 04 '20 at 01:14
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It seems you haven't read the liunked post, and I am not going to repeat it in comments here. If you do so then the point of my initial comment should be clear. – Bill Dubuque Apr 04 '20 at 01:21
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@ Bill Dubuque. The generality of the remainder theorem you mentioned earlier (applying to all rings) is false. I don't have to read a link to tell you that. The point is that the remainder theorem applies here as $\Bbb F_7$ is a field. From that we have the factor theorem. The rest for multiple factors is an obvious consequence of the remainder theorem again and as I pointed out at length above. – Oliver Kayende Apr 04 '20 at 01:21
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I see nothing false in the linked posts. It seems you are missing my point. You wrote "implying" without justifying why that implication is true. That implication depends crucially on the coef ring being a field (or domain), so you should at least mention that if you are not going to justify the implication. It is not just nitpicking since many students make errors exactly at this point when they wander into more general rings. As such, we should strive to give rigorous arguments in order to help them avoid making these mistakes. – Bill Dubuque Apr 04 '20 at 01:27
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@Bill Dubuque. You wrote "You wrote "implying" without justifying why that implication is true, That implication depends crucially on the coef ring being a field (or domain)"...I concurred and wrote just this in my first comment response. You wrote "many students make errors exactly at this point...we should strive...to help them..."... I deeply concur with you again on this point and I hope that this exchange has fettered that rigorous argument out for all concerned. – Oliver Kayende Apr 04 '20 at 01:44
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@Bill Dubuque. You wrote "And it's not the Polynomial Remainder Theorem that fails (that's true over any ring)"...I believe this is false. Consider $f(x)=7x^2+2x-1,g(x)=3x+2\in\Bbb Z[x]$ and try to apply the remainder theorem only to fail. – Oliver Kayende Apr 04 '20 at 01:45
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Almost universally, PRT refers to the case of a monic linear divisor, and that case does indeed hold true over any commutative ring - as I wrote. Same for division by any monic divisor (in fact there is a simple generalization of the Polynomial Division Algorithm to nonmonic divisors as I explain here). If you had followed the links this would have been clear. – Bill Dubuque Apr 04 '20 at 02:01
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Both polynomials of degree 6 and leading coefficient 1 have the same 6 zeroes in the prime field $\mathbb F_7$.
miracle173
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$\;\;\;$ Generalize from $\Bbb F_7=\frac{\Bbb Z}{7\Bbb Z}$ to any prime finite field $\Bbb F_p=\frac{\Bbb Z}{p\Bbb Z}$ and use only non-constant monic polynomials with coefficients in $\Bbb F_p$. By Lagrange's theorem we must have $f(\alpha)=\mathbf 0$ for each $\alpha$ in the multiplicative group $\Bbb F_p^*:=\{\mathbf{1,\dots,p-1}\}$ where $$f(x):=x^{p-1}-\mathbf 1\;\;\text{and}\;\;g(x):=(x-\mathbf 1)\cdots(x-\mathbf (p-1))=\prod_{\alpha\neq\mathbf 0}(x-\alpha)$$ which implies the polynomial $h(x):=f(x)-g(x)$ has at least $p-1$ distinct roots but at most degree $p-2$.
$\;\;\;$ Therefore $f(x)-g(x)=\mathbf 0$ or else, of the non-constant polynomials each with more distinct roots than its polynomial degree, we could choose monic $F(x)$ of least degree but then by the Polynomial Factor Theorem $F(x)=(x-\alpha)\cdot G(x)$ for some root $\alpha$ and because $\Bbb F_p$ is a field the remaining distinct roots of $F(x)$ must be roots of $G(x)$ which would imply $G(x)$ has more distinct roots than its degree which is impossible because with this property $F(x)$ is the monic polynomial of least degree.
Oliver Kayende
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