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Given two polynomials $A = \sum_{0\le k<n} a_k x^k$ and $B =\sum_{0\le k<n} b_k x^k$ of the same degree $n$, which are equal for all $x$, is it always true that $\ a_k = b_k\ $ for all $0\le k<n?$. All Coefficients and $x$ are complex numbers.

Edit: Sorry, formulated the question wrong.

FUZxxl
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2 Answers2

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The answer is in general no. If the ground field is infinite,then it is true. In general it is not TRUE. In the polynomial algebra ${\mathbb{Z}/2\mathbb{Z}}[X]$ consider the polynomials $X^2$ and $X$. But they are different in ${\mathbb{Z}/2\mathbb{Z}}[X]$.

A.G
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  • A=B for all x is the premise, so how can A!=B? – TROLLHUNTER Mar 04 '11 at 15:21
  • @kakemonsteret: The distinction is between the two questions "When are two polynomials equal?" and "when are two polynomial functions equal?". The answer to the second is "whenever A(x) = B(x) for all x" while the answer to the first is "whenever A and B have the same coefficients."

    Is it easy to see that if the ground field is infinite, the two questions coincide?

     – Jason DeVito - on hiatus Mar 04 '11 at 15:24
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    @kakemonsteret: take $A = x, B = x^p$ over the ground field $k = \mathbb{F}_p$. Then $A(t) = B(t)$ for all $t \in k$ by Fermat's little theorem, but $A \neq B$ in $k[x]$. In other words, over finite fields polynomials cannot be identified with the functions they induce. – Qiaochu Yuan Mar 04 '11 at 15:25
  • Sorry, formulated the question wrong. – FUZxxl Mar 04 '11 at 15:39
  • This answer still applies to your new formulation. – Qiaochu Yuan Mar 04 '11 at 15:59
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    It seems to me he was just asking about "ordinary" polynomials one would encounter in high school math, i.e. over R. – Harry Stern Mar 04 '11 at 16:14
  • @Harry Stern: Yes :) – FUZxxl Mar 05 '11 at 08:21
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For $\rm\ f = A-B\in R[x]\:,\:$ it is equivalent to ask if $\rm\ f(r) = 0\ $ for all $\rm\: r\in R\ \Rightarrow\ f = 0\:,\: $ i.e. if $\rm\:f\ $ is zero as a function then is $\rm\:f\ $ zero as a formal polynomial, i.e. are all its coefficients zero? This is true if $\rm\:R\:$ is an integral domain of cardinality greater than the degree of $\rm\:f\:,\:$ e.g. if $\rm|R|$ is infinite, but it may fail otherwise, e.g. $\rm\ x^p = x\ $ for all $\rm\: x\in \mathbb Z/p\ $ by Fermat's little theorem, but $\rm\ x^p \ne x\ $ in $\rm\: \mathbb Z/p\:[x]\:.$

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 showing that, $\:$ given any $\rm\:f(x)\:$ with more roots than its degree,$\:$ we can quickly compute a nontrivial factor of $\rm\:m\:$ via 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 $1$ that is not $\:\pm 1$.

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Bill Dubuque
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