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The book I'm reading introduces polynomials over a field and proves the statement that a polynomial of degree $n$ has at most $n$ zeros. They do this by using division algorithm and induction.

Then they make the following remark:

This is not true for polynomial over arbitrary rings. For instance $x^2 + 7 \in \mathbb{Z}_8$ has roots $1,3,5,$ and $7$.

My question: what fails in the previous proof for arbitrary rings? They just made that remark and moved on.

Edit -- Proof (from Gallian):

We proceed by induction on $n$. Clearly, a polynomial of degree $0$ over a field has no zeros. Now suppose that $f(x)$ is a polynomial of degree $n$ over a field and $a$ is a zero of $f(x)$ of multi­plicity $k$. Then, $f(x)=(x-a)^kq(x)$ and $q(a) \neq 0$. Note we have $\text{deg }f = n = k + \text{deg }q$. If $f(x)$ has no zeros other than $a$, we are done. On the other hand, if $b \neq a$ and $b$ is a zero of $f(x)$, then $0=f(b)=(b-a)^kq(b)$ so that $b$ is a zero for $q(x)$ with the same multiplicity it has for $f(x)$. By the Second Principle of Mathematical Induction, we know that $q(x)$ has at most deg $q(x)=n-k$ zeros, counting multiplicity. Thus, $f(x)$ has at most $k + n -k = n$ zeros, counting multiplicity.

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  • $\begingroup$ Can you sketch the proof you are referring to? $\endgroup$ – Florian R Jun 9 '18 at 14:55
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    $\begingroup$ Why don't you go through the proof line by line and see what hypotheses are used and whether they hold in $\mathbb{Z}/8\mathbb{Z}$? $\endgroup$ – Alex B. Jun 9 '18 at 14:55
  • $\begingroup$ well I have a hunch, I'm not sure though. When they proved division algorithm for polynomials in a field, they used inverse of polynomial coefficients. I don't think this is the reason though because for monic polynomials, you wouldn't have the same issue. (I realize I haven't put the proofs in, I'm doing that now) $\endgroup$ – yoshi Jun 9 '18 at 15:01
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Usually, you can go from a factorization like $(x-1)(x-2)=0$ to the pair of equations $x-1=0$ and $x-2=0$ to get your solutions. This uses the zero product property, which says that if $AB=0$ then either $A$ or $B$ are equal to zero. This fails over arbitrary rings. For example, in $\mathbb{Z}/8\mathbb{Z}$, the zero product property fails because $2\cdot 4=4\cdot 4=4\cdot 6=0$. So if we want to solve $(x-1)(x-7)=0$, then there are six things that can happen. Either: $$ (1)\;\;\;\;x-1 = 0\;\;\textrm{or}\;\; x-7=0\\ (2)\;\;\;\;x-1 = 2\;\;\textrm{and}\;\; x-7=4\\ (3)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=2\\ (4)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=4\\ (5)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=6\\ (6)\;\;\;\;x-1 = 6\;\;\textrm{and}\;\; x-7=4 $$ The only ones that can happen are (1), (2) and (5) using $x=1,7$ and $x=3$ and $x=5$ respectively. Therefore, the solutions are $1,7,3,5$. Note that the extra zeros correspond both to zero divisors of the base-ring, as well as two different factorizations in the polynomial ring.

In general, if $R$ has zero divisors, then $R[x]$ will not have unique factorization since there is a correspondence between zeros and factors of a polynomial. The zero divisors can ensure that there are "too many" factors.

For instance, let $a,b$ be zero divisors and let $c,d$ be any other values so that $a-b=c-d$. Set $s=a+d=b+c$. Consider the polynomial $p(x)=(x-c)(x-d)$. Clearly $x=c,d$ are roots. But, furthermore, $p(s)=(b+c-c)(a+d-d)=ab=0$. So $x=s$ is also a root, and you can check that $x=t$, where $t=c+d-s$, is also a root. We then get $p(x)=(x-c)(x-d)=(x-s)(x-t)$.

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The issue is that the polynomial division is not unique in $\mathbb Z_8$. This is the consequence of existence of zero divisors.

You’ll be able to verify that $$x^2+7=(x-1)(x-7)=(x-3)(x-5)$$

providing evidence that $1, 3, 5,7$ are roots.

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  • $\begingroup$ Nailed it. This is exactly what this is about: zero divisors. Not multiplicative inverses. A polynomial $f$ over an integral domain always has at most $\deg(f)$ roots. $\endgroup$ – freakish Jun 9 '18 at 16:10
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    $\begingroup$ In the case of noncommutative rings, it’s not just about zero divisors. $x^2+1=0$ is satisfied by infinitely many elements of $\mathbb H$, for example. (Probably the OP had commutative tings in mind, but this is worth knowing.) $\endgroup$ – rschwieb Jun 9 '18 at 16:23
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The problem is the existence of zeros divisors in the ring.

Polynomial division does not always work over rings that are not fields. For instance, $x^2-5$ cannot be divided by $2x+1$ over $\mathbb Z$.

Polynomial division by monic polynomials does work over all rings. So, if $a$ is a root of a polynomial $p$ over a ring $R$, then $p(x)=(x-a)q(x)$. If $b\ne a$ is another root of $p$, then $(b-a)q(b)=0$. Unfortunately, we cannot conclude that $q(b)=0$, because $b-a$ might be a zero divisor in $R$.

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