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This is from Pinter, A Book of Abstract Algebra, p.265.

Given $p(x) \in F[x]$ where $F$ is a field, I would like to show that $p(x)$ divided by $x-c$ has remainder $p(c)$.

This is easy if $c$ is a root of $p$, but I don't see how to prove it if $c$ is not a root.

Bill Dubuque
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user9352
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5 Answers5

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Division Algorithm $\rm\:\!\Rightarrow p(x) = q(x)\ (x\!-\!c) + r,\ r\in F.\, $ Eval at $\rm\ x\! =\! c\ \Rightarrow\ p(r)=r.\ \bf\small QED$

Therefore $\rm\ x - c\ |\ p(x)-p(c)\ $ in $\rm\:R[x],\,$ $\rm \ \forall\ p\in R[x]\:,\ \forall\ rings\ R\quad $ [Factor Theorem]

Said equivalently $\rm\ p(x)\equiv p(c)\ \pmod{\!x - c}$

or, in other words, $\rm\ x\equiv c \,\Rightarrow\, p(x)\equiv p(c)\,\ $ [also follows by Polynomial Congruence Rule]

E.g. casting nines: mod $\rm 9\!:\ 10\equiv 1\ \Rightarrow\ N= p(10)\equiv p(1) \equiv $ sum of digits of $\rm\, N\,$ in radix $10$.

Note how the result is much clearer in the language of congruence arithmetic.

Remark $\ $ The shift automorphism $\rm\ x\to x+c\ $ reduces the proof of the Factor Theorem to the "obvious" special case $\rm\:c = 0,\:$ e.g. see here and my sci.math post appended below. However, this approach is a bit risky pedagogically since such a proof is not completely rigorous without knowledge that such maps are ring automorphisms. It is essential that students learn how to make rigorous (ring-theoretically!) prior informal arguments about substitution, changing "variables", etc. Much subtlety lies here, e.g. even in the informal notation for polynomials, such as $\rm\: P(X\!+\!c)\:$ below. This automorphism is the essence behind Patrick's answer, and is also implicitly in Pierre's. Of course such shifting is yet another example of transformation-based problem-solving, cf. my recent post on analogously applying a shift so that Eisenstein's irreducibility criterion applies.

asdf [email protected] wrote to sci.math on 29 Mar 2006 (paraphrased)

How do you prove that for a polynomial P(X)

$\rm P(c)=0\ \Rightarrow\ X-c\ |\ P(X)\ $ i.e. $\rm\ (X-c)\ Q(X)\ =\ P(X)\ $ for some $\rm\ Q(X)\ $

For $\rm\ c=0\ $ it specializes to the obvious case: $\rm\ X\mid P(X) \iff P(0)=0$

If $\rm\ c\ne 0\ $ reduce to $\rm\ c=0\ $ by a shift: $\rm\ X\!-\!c\mid P(X) \iff X\mid P(X\!+\!c) \iff P(c)=0 $

It is helpful to be aware of the following simple equivalences.

Theorem $\ $ TFAE for a polynomial $\,f\in R[x],\,$ and $\,a\in R\,$ a commutative ring.

$(0)\ \ \ f = (x\!-\!a)q + r\ $ for some $\,q\in R[x],\ r\in R\ \ \ $ [Monic Linear Division Algorithm]

$(1)\ \ \ f\bmod (x\!-\!a) = f(a)\ \ \ \ $ [Remainder Theorem]

$(2)\ \ \ f(a) = 0\,\Rightarrow\, x\!-\!a\mid f\ \ \ \ $ [Factor Theorem]

Proof $\ (0\Rightarrow 1)\ \ \ f = (x\!-\!a)q + r\,\overset{\large x\,=\,a}\Longrightarrow\, r=f(a)\,\Rightarrow\,f\bmod x\!-\!a = r = f(a) $

$(1\Rightarrow 2)\ \ \ f\bmod x\!-\!a = f(a) = 0\,\Rightarrow\, x\!-\!a\mid f$

$(2\Rightarrow 0)\ \ \ g := f-f(a)\,$ has $\,g(a) = 0\ $ so $\ g = f-f(a) = (x\!-\!a)q$

Remark $ $ The polynomial division algorithm always works in any polynomial ring for divisors that are monic (lead coef $= 1$ or a unit) such as $\,x-a\,$ above, see here.

Bill Dubuque
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By the division algorithm, if $a(x)$ and $b(x)$ are any polynomials, and $a(x)\neq 0$, then there exist unique $q(x)$ and $r(x)$ such that $$b(x) = q(x)a(x) + r(x),\qquad r(x)=0\text{ or }\deg(r)\lt \deg(a).$$

Let $b(x) = p(x)$, and $a(x)=x-c$. Then $r(x)$ must be constant (since it is either zero or of degree strictly smaller than one), so $$b(x) = q(x)(x-c) + r.$$ Now evaluate at $x=c$.

Note. I find it strange that you say that this is "easy if $c$ is a root of $p(x)$". The Factor Theorem (that $x-c$ divides $p(x)$ when $c$ is a root of $p(x)$) is a corollary of this result. How exactly do you prove it without this?

Arturo Magidin
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  • Is it required to say that $F[x]$ is an Euclidean Domain for any field $F$? –  Dec 28 '11 at 17:19
  • Perhaps you could go on and prove that? Hint : use induction on the degree of the polynomial you divide. – Patrick Da Silva Dec 28 '11 at 17:21
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    @KannappanSampath: "Required" by whom? It all depends on what background material you are allowed to assume and what you are not. – Arturo Magidin Dec 28 '11 at 17:23
  • Actually this result is equivalent to the Factor Theorem, since $P(x)-P(c)$ always has $c$ as a root. Anyhow, I agree with you, usually the Factor Theorem is proven by proving this result first. – N. S. Dec 28 '11 at 17:44
  • @Arturo well I knew this result it is basically what he does in the book but I guess what I was missing was just plug c in for x – user9352 Dec 28 '11 at 17:44
  • @Kannappan The assumption that the Divison Algorithm holds in $F[x]$ is equivalent to assuming that $F[x]$ is a Euclidean domain. The former simply uses more elementary language, so it will be accessible to readers who have not yet studied the abstract concept of Euclidean domains. – Bill Dubuque Dec 28 '11 at 18:00
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Here's how it goes : the polynomials $\{1, (x-c), (x-c)^2, \dots \}$ form a basis of the vector space $F[x]$. Write $$ p(x) = a_0 + a_1 (x-c) + a_2 (x-c)^2 + \dots + a_n (x-c)^n. $$ Then $$ p(x) = (x-c) \left( a_1 + a_2(x-c)^2 + \dots + a_n (x-c)^{n-1} \right)+ a_0 $$ and you can see that $p(c) = a_0$.

Hope that helps,

Patrick Da Silva
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  • As Bill Dubuque says, division algorithm works too, I just thought I'd be more explicit. Actually this also provides one way to show the Euclidean algorithm works (for the polynomial $(x-c)$ in particular) : write $p(x) = p((x-c) + c)$ and in the expression of $p(x)$ as $\sum_{k=0}^n a_k x^k$, you replace $x^k$ by $(x-c+c)^k$ and expand this by using the binomial theorem and keeping the $(x-c)$ terms together. – Patrick Da Silva Dec 28 '11 at 17:11
  • While that works in this special case, it doesn't generalize to higher degree divisors. It's really the division algorithm that is the essence of the matter here, i.e. that $\rm:F[x]:$ is a Euclidean domain. More generally, in any polynomial ring the polynomial long division algorithm works for any divisor whose leading coefficient is a unit, i.e. invertible. – Bill Dubuque Dec 28 '11 at 17:47
  • I know it doesn't for higher degree polynomials, but this argument is often forgotten and sometimes more intuitive than the Euclidean algorithm. Being a fan of intuition myself I like to recall this argument every once in a while. – Patrick Da Silva Dec 28 '11 at 18:25
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    One can also do this ring theoretically by exploiting the shift automorphism $\rm\ x\to x+c:,:$ see my remark in my answer. – Bill Dubuque Dec 28 '11 at 20:40
  • One needs to be very careful that such a proof isn't circular. To prove that $\rm:{:(x-c)^{:i:}}:$ spans $\rm:F[x]:$ one very natural approach is to repeatedly apply the Division Algorithm to obtain the Horner form $\rm:p(x)\ =\ a_0 + (x-c)(a_1 + (x-c)(a_2+(x-c)(a_3+:\cdots:))):.:$ Did you have in mind some other method that is not so blatantly circular? – Bill Dubuque Dec 28 '11 at 21:37
  • Automorphisms sends a basis to another, do they? If $\phi$ maps $x$ to $x-c$ then it sends the canonical basis to the other one. I don't think you need a proof to show that the powers of $x$ are a basis of $F[x]$ since we literally define $F[x]$ as linear combinations of powers of $x$. Proving that $\phi$ is an automorphism is fairly trivial and I don't think it relies on the division algorithm. I think you're just worrying too much about technicalities. If you don't wanna use the automorphism method, robjohn's technique works too by using induction on the degree. – Patrick Da Silva Dec 29 '11 at 00:04
  • I didn't realize that you were (implicitly) invoking the fact that the shift $\rm\ x\to x+c\ $ is an automorphism of $\rm:F[x]:.:$ Perhaps you should explicitly mention that so it is clear what is intended. As for technicalities, they are essential in mathematics. – Bill Dubuque Dec 29 '11 at 00:24
  • I don't need this at all, there are plenty of ways to do this, one is to show that $x^k = x^k-c^k + c^k = (x-c)(polynomial) + c^k$ and use induction, another one would be $$ \sum_{k=0}^n a_k x^k = \sum_{k=0}^n a_k (x-c+c)^k = \sum_{k=0}^n a_k \left( \sum_{i=0}^k \begin{pmatrix} n \ i \end{pmatrix} (x-c)^i c^{n-i} \right) $$ and you have your basis right there. Like I said, you're worrying too much about technicalities. If you are still worried, I can't do much. – Patrick Da Silva Dec 29 '11 at 00:33
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    Of course there are many ways to do it - some correct, some circular. But, alas, I have no clue from what you wrote in your answer which alternative applies. Hence my remarks. Good mathematical exposition does not leave such doubts in the mind of the reader. It would be nice it you elaborated in your answer precisely how you propose to deduce said elements form a basis. – Bill Dubuque Dec 29 '11 at 00:55
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Suppose that $$ p(x)=\sum_{k=0}^na_kx^k $$ Then $$ p(x)-p(c)=\sum_{k=0}^na_k(x^k-c^k) $$ For each $k$, we have that $$ \frac{x^k-c^k}{x-c}=x^{k-1}+x^{k-2}c+x^{k-3}c^2+\dots+xc^{k-2}+c^{k-1} $$ Thus, $$ \begin{align} \frac{p(x)-p(c)}{x-c} &=\sum_{k=0}^na_k(x^{k-1}+x^{k-2}c+x^{k-3}c^2+\dots+xc^{k-2}+c^{k-1})\\ &=q(x) \end{align} $$ Therefore, $p(x)=q(x)(x-c)+p(c)$, which is another way of writing $p(x)$ divided by $x-c$ leaves a remainder of $p(c)$ since the degree of $p(c)$, $0$, is less than the degree of $x-c$, $1$.

robjohn
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  • This is essentially a different way of handling the modular equivalences in Bill Dubuque's answer. – robjohn Dec 28 '11 at 19:24
  • Yes, by linearity, $\rm\ x-c\ |\ p(x)-p(c)\ $ reduces to the well-known monomial case $\rm\ p(x) = x^k:.:$ In terms of congruences $\rm\ x:\equiv:c\ \Rightarrow\ x^k:\equiv:c^k\ \Rightarrow\ \sum a_k x^k:\equiv \sum a_k c^k:$ $:\Rightarrow:$ $\rm:p(x):\equiv:p(c):.$ – Bill Dubuque Dec 28 '11 at 20:54
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"$p(x)$ divided by $x-c$ has remainder $p(c)$"

is equivalent to

"$p(x+c)$ divided by $x$ has remainder $p(c)$",

which is equivalent to

"$q(x)$ divided by $x$ has remainder $q(0)$",

which is obvious.

EDIT. This is essentially the argument used By Gauss in Article 43 of the Disquisitiones Arithmeticae. A French translation is available at Internet Archive and at Google Books:

Pierre-Yves Gaillard
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