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I was reading in a book that the Factor Theorem only holds over fields (not rings). Why would that be true? No where in the proof of the factor theorem is a multiplicative inverse taken, so the proof should be as valid for polynomials in rings as it is for polynomials in fields, right?

user26857
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user1299784
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    Depends on how you state the factor theorem. Over a general commutative ring, $x-a$ is a factor of $p(x)$ if and only if $p(a)=0$. But there can be more than $n=\deg(p)$ distinct roots. – Thomas Andrews Jan 21 '14 at 04:55

2 Answers2

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The single factor theorem holds over any commutative ring. But multiple factors need not, e.g.

Bifactor Theorem $\ $ Suppose that $\rm\,a,b\,$ are elements of a commutative ring $\rm\,R\,$ and $\rm\:f\in R[x]\,$ is a polynomial with coefficient in $\rm\,R.\,$ If $\rm\ \color{#C00}{a\!-\!b}\ $ is cancelable in $\rm\,R\,$ then

$$\rm f(a) = 0 = f(b)\ \iff\ f\, =\, (x\!-\!a)(x\!-\!b)\ h\ \ for\ \ some\ \ h\in R[x]$$

Proof $\,\ (\Leftarrow)\,$ clear. $\ (\Rightarrow)\ $ Applying Factor Theorem twice, while canceling $\rm\: \color{#C00}{a\!-\!b},$

$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in R[x]\\ \rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in R[x]\\ &\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$

Beware $\ $ The theorem may fail when $\rm\ a\!-\!b\ $ is not cancelable (i.e. is a zero-divisor), e.g.

$\rm mod\ 8\!:\,\ f(x)=x^2\!-1\,\Rightarrow\,f(3)\equiv 0\equiv f(1)\,$ $\rm but\ \ x^2\!-1\not\equiv (x\!-\!3)(x\!-\!1)\equiv x^2\!-4x+3$

Bill Dubuque
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    (+1) I do think that you need $R$ to be commutative. Otherwise the implication $f(x)=(x-b)g(x)\implies f(a)=(a-b)g(a)$ may not hold. As an example consider $x^2+1$ over the quaternions. – Jyrki Lahtonen Jan 21 '14 at 05:18
  • The book says on page 146 of the linked PDF: "All the results concerning polynomials over fields (eg. factor theorem) hold over Zp, the set of integers modulo p, [if p is prime]. This is not the case if p is not prime. For example, 2x^3-2x has 4 zeroes in the ring (not field) Z4 (0, 1, 2, and 3) but only 3 in Z. – user1299784 Jan 21 '14 at 06:26
  • @user1299784 The point is that many familiar results from field theory no longer hold true over more general rings, because their proof may depend on properties that no longer hold true, such as nonzero elements are invertible or cancelable. For example, over a field the Bifactor Theorem is usually stated with the hypothesis that the roots $,a,b,$ are distinct (vs. $,a-b,$ is cancellable). But this field-theoretic form of the Bifactor Theorem is false for more general rings, as the example in the Remark shows. – Bill Dubuque Jan 21 '14 at 06:55
  • I quite agree that in the context of polynomial rings it is most common to assume that $R$ is commutative, and that may be tacitly assummed. My remark originated from the context that the OP is explicitly seeking to generalize facts from fields to ... whatever rings. Therefore it felt ok as a parenthetical remark to mention that commutativity is necessary. I apologize if my remark sounded like a piece of criticism as opposed to a parenthetical comment. I did upvote both the answers :-) – Jyrki Lahtonen Jan 21 '14 at 08:33
  • Note on notation Generally I assume ring denotes "commutative ring" on general-level math sites like this see this meta discussion. But I do agree that it is worth emphasis that certain common results do require commutativity. – Bill Dubuque May 12 '20 at 16:36
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What theorem are you calling the "factor theorem" ? If you mean this result:

(1) If $a$ is a root of a polynomial $f(x)$, then $f(x) = (x-a)q(x)$ for some polynomial $q(x)$.

It is true that (1) continues to hold over arbitrary rings. However, some standard consequences of (1) no longer hold over arbitrary rings. For example (to take the example from your link),

(2) If $f(x)$ is a polynomial of degree $n$, then $f(x)$ has at most $n$ roots.

This is false over general rings, as the example in your link shows. The usual proof of (2) is to take a root $a$ and then use (1) to write $f(x) = (x-a)q(x)$. The next step is to note that if $c$ is any other root of $f(x)$ besides $a$, then $c$ is a root of $q(x)$. This step fails in a general ring. For if we substitute $x=c$ we get $0 = f(c) = (c-a)q(c)$, and in a general ring we cannot cancel the $c-a$ term (even though we know it is nonzero) to conclude that $q(c)=0$.

Ted
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    (+1) I do think that you need $R$ to be commutative. Otherwise the implication $f(x)=(x-a)q(x)\implies f(c)=(c-a)q(c)$ may not hold. As an example consider $x^2+1$ over the quaternions. – Jyrki Lahtonen Jan 21 '14 at 05:21
  • The book says on page 146 of the linked PDF: "All the results concerning polynomials over fields (eg. factor theorem) hold over $Z_p$, the set of integers modulo p, [if p is prime]. This is not the case if p is not prime. For example, 2x^3-2x has 4 zeroes in the ring (not field) $Z_4$ (0, 1, 2, and 3) but only 3 in $Z$. – user1299784 Jan 21 '14 at 06:17
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    @JyrkiLahtonen For the single factor theorem we don't need $R$ commutative: see Lam, A First Course in Noncommutative Rings, 2001, Proposition 16.2. – user26857 Oct 06 '16 at 08:54