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In considering: A polynomial over a field of degree n has at most n roots.

-- How does this make use of the stipulation "over a field" - especially with an eye toward inveritbility ? Is it to insure one can obtain a monic polynomial? Or perhaps because the Euclidean algorithm might apply in the proof?

Not unrelated, in looking for an intuitive understanding:

-- In $\mathbb{Z}/8\mathbb{Z}$[$x$], which is not a field, $x^2$-1 has degree 2, and has 4 roots {1,3,5,7}. While, e.g., the same polynomial has two roots in $\mathbb{Z}/7\mathbb{Z}$[$x$], over a field.

Here my question is, how does not being or being over a field distinguish these two results. Especially from the perspective that every element of a field has a multiplicative inverse.

And as a follow-up, it seems a bit "ironic" (probably not a word applicable to math) that in the latter case all the elements are prime to 7 (invertible, in a field) yet there are only two roots (= degree of the polynomial). Yet in the first case, the four roots are themselves elements that are prime to 8 (invertible).

So here I would appreciate an intuitive understanding of how invertibility applies in these two cases. How does it makes things work in the case of mod 7 to satisfy the above theorem regarding number of roots and the degree of the polynomial. And how in the case of mod 8, the theorem is not applicable, and further, that the roots are in fact the very elements that are invertible mod 8.

As usual, I am a self-studier, so please forgive me if these questions are naive or poorly phrased. Thanks for your patience and kind reply.

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    @DimitrijeKostic - no $\mathbb Z/p^n\mathbb Z$ is not a field when $n>1$. There exists a finite field of order $p^n$ for all $n$, but that ring is not it. – Thomas Andrews Dec 22 '11 at 19:02
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    @Andrew: The "induction on the degree" that you comment on below does not work, when you have zero divisors. As Bill Dubuque says, over $\mathbf{Z}/8\mathbf{Z}$ you have $x^2-1=(x-1)(x+1)$. Yet $x=3$ is a zero, because $(3-1)(3+1)=2\cdot4=8=0$ in this ring. The failure comes from the fact that non-zero elements can have product zero. So even if $f(x)=(x-r)g(x)$ the polynomial $f(x)$ may have zeros that are neither equal to $r$ nor zeros of $g(x)$. – Jyrki Lahtonen Dec 22 '11 at 20:11
  • @JyrkiLahtonenThanks, that was exactly what I was wondering about. I appreciate your tying those two aspects of my question together. (Still working on the other answers.) –  Dec 22 '11 at 20:50

3 Answers3

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The real issue (for commutative rings) is zero divisors.

Suppose $R$ is an integral domain (commutative ring with $1 \not= 0$ and no zero divisors). Then $R$ can be embedded in its field of fractions $R \subseteq \mathbb{F}$. So if $f(x) \in R[x]$, then $f(x) \in \mathbb{F}[x]$. Thus if $f$ has degree $n$, it has at most $n$ roots.

The real reason the "at most $n$ roots" part holds is because integral domains obey cancellation laws. So if you have a root $r$ so that $f(x)=(x-r)g(x)$ then you can cancel off $(x-r)$ and are left with $g(x)$ which is a polynomial of lower degree (then continue using induction).

If $R$ is not an integral domain, then cancellation laws do not hold and so you can have more than $n$ roots for a polynomial of degree $n$.

Now if $R$ isn't even commutative (even if $R$ has no zero divisors) it gets worse. Polynomials over the quaternions (a "non-commutative" field -- i.e. a skew field or division ring) can have more roots than their degrees.

So in general, if $R$ is non-commutative or if $R$ is commutative but with zero divisors, then polynomials with coefficients in $R$ do not necessarily have unique factorizations and so polynomials of degree $n$ can have more than $n$ roots.

Edit Quaternion example: Let $\mathbb{H} = \{ a+bi+cj+dk \;|\; a,b,c,d\in \mathbb{R} \}$ be the (real) quaternions. Then the polynomial $x^2+1 \in \mathbb{H}[x]$ has more than $2$ roots. In fact, $\pm i, \pm j, \pm k$ are all roots of $x^2+1$ (since $i^2=j^2=k^2=-1$).

Bill Cook
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    Also worth noting that if $r$ is a root of $f(x)$ then you can always write $f(x)=(x-r)g(x)$. $R[x]$ doesn't have a division algorithm in general, but does when dividing by monics, and, particularly forthe case $x-r$. – Thomas Andrews Dec 22 '11 at 19:23
  • @ThomasAndrews Thanks. I am studying all the answers. So maybe this follow-up is dealt with, but based solely on your above remark, why does the theorem stipulate a "polynomial over a field"? Using f(x) = (x - r) g(x), I think you can show the number of roots < or = to the degree n of f using induction, assuming g has < or = (n-1) roots. So is it that "over a field" is needed just to enable division of the coefficient to get a monic? Thanks. –  Dec 22 '11 at 20:02
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    It may be worth noting that in the quaternion case the polynomial $x^2+1$ has infinitely many roots. All the quaternions with $a=0, b^2+c^2+d^2=1$ are zeros. Even though there are no zero divisors. All because factorization in a ring like $\mathbf{H}[x]$ doesn't really make sense. At least not for the purposes of finding roots, because the powers of the unknown $x$ are multiplied from the left and from the right, and in a non-commutative ring that leads to madness. Evaluating a polynomial with real coefficients at a quaternion makes sense, because $\mathbf{R}$ is the center of $\mathbf{H}$. – Jyrki Lahtonen Dec 22 '11 at 20:05
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    @Andrew Yes most likely the theorem states the result for polynomials with field coefficients, just to simplify the discussion. One could say, "The division algorithm holds over any commutative ring (with $1$) as long as the polynomial you are dividing by has a leading coefficient which is a unit." But it's simpler to say "You can do polynomial long division dividing by any non-zero polynomial when working with field coefficients." – Bill Cook Dec 22 '11 at 20:06
  • @BillCook Thanks –  Dec 22 '11 at 20:37
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Here is one insightful way to view it. Consider the following proof for an integral domain $\rm\,R.$

$\rm\ \ a_i\ne a_j\ \Rightarrow\ \{x-a_i\}\ $ are nonassocate primes in $\rm\:R[x],\:$ by $\rm\: R[x]/(x\!-\!a) \cong R\:$ is a domain.

Therefore $\rm\ \ x\!-\!a_1\ |\ f(x),\ \ldots\:,\: x\!-\!a_n\ |\ f(x)\ \ \Rightarrow\ \ (x\!-\!a_1)\cdots (x\!-\!a_n)\ |\ f(x) $

since LCM = product for nonassociate primes. But this is contra degree if $\rm\ n > deg\ f.$

When $\rm\:R\:$ is not a domain this argument fails since then $\rm\: x\! -\! c\:$ is no longer prime, e.g. $\rm\ (x\!-\!1)\, (x\!+\!1)\, =\, (x\!-\!3)\, (x\!+\!3)\ $ over $\rm\:\mathbb Z/8,\:$ so none of the factors are prime.

Remark $\ $ In fact a ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in $\rm\:D.\:$ For the simple proof see my post here, where I illustrate it constructively in $\rm\: \mathbb Z/m\: $ by proving

$\quad$ if $\rm\:f(x)\in\mathbb Z/m\:$ has more roots than its degree,$\:$ we can quickly factor of $\rm\:m\:$ by a $\rm\:gcd\:$

The quadratic case of this result is at the heart of many integer factorization algorithms, which try to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m,\:$ e.g. a square root of $1$ not $\,\pm 1$.

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Bill Dubuque
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It's not clear to me why you think it is ironic.

So let's look at the case you've given: $x^2-1 = (x-1)(x+1)$.

Note that in a field, if $x^2-1=0$ and $x+1\neq 0$, you can multiply both sides by $\mathbb (x+1)^{-1}$ and see that $x-1 = 0$.

What you really need to be able to say is that the ring has no zero divisors.

In finite rings, no zero divisors is the same as inverses always existing - no zero divisors means that for any $a\neq 0$, $x\to ax$ is a $1-1$ map, and hence, if your ring is finite, it is necessarily onto, so $ax=1$ for some $x$.

The other feature of a field you need is commutativity - for example, in quaternions, there are more than two roots to $x^2+1$.

So, in a commutative ring without zero divisors, any polynomial of degree $n$ can have at most $n$ roots.

Thomas Andrews
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  • @ThomasAndrewsAlthough probably pretty silly (so forgive me); but here's what I think is ironic. Suppose you "want" the theorem to hold, then you need for mod n to be mod a prime - then all the elements are prime. So you are hoping for all primes. But in the case of mod 8, say, while you are hoping for primes, it is the prime elements themselves that give you more roots and take you over the top (the degree of the polynomial). –  Dec 22 '11 at 21:24
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    No, in any ring you can divide by monics, it's just that once you have a factorization of the form: $p(x)=(x-a_1)...(x-a_n)$, if there are zero divisors in your field, it is possible for $p(x)=0$ with $x\notin{a_1,...,a_n}$ – Thomas Andrews Dec 22 '11 at 21:58