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.

  • $\begingroup$ Yes. Consider g(x) = A - B. $\endgroup$ – Harry Stern Mar 4 '11 at 15:11
  • $\begingroup$ You just said A=B for all x. x is the only variable, so yes. $\endgroup$ – TROLLHUNTER Mar 4 '11 at 15:13
  • $\begingroup$ As Anjan mentions below, it depends on what kind of things the $a_k$ and $b_k$ are. Would you mind clarifying? $\endgroup$ – Jason DeVito Mar 4 '11 at 15:21
  • $\begingroup$ @Jason DeVito: $a_k$ and $b_k$ are just coefficients. They're independent of $x$. $\endgroup$ – FUZxxl Mar 4 '11 at 15:37
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    $\begingroup$ That's not the question. The question is what field or ring they lie in. Are they integers? Real numbers? Elements of a finite field? $\endgroup$ – Qiaochu Yuan Mar 4 '11 at 15:59

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 $\:\pm 1$.


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 ${\mathbb{Z}/2\mathbb{Z}}[X]$.

  • $\begingroup$ A=B for all x is the premise, so how can A!=B? $\endgroup$ – TROLLHUNTER Mar 4 '11 at 15:21
  • $\begingroup$ @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? $\endgroup$ – Jason DeVito Mar 4 '11 at 15:24
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    $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Mar 4 '11 at 15:25
  • $\begingroup$ Sorry, formulated the question wrong. $\endgroup$ – FUZxxl Mar 4 '11 at 15:39
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    $\begingroup$ It seems to me he was just asking about "ordinary" polynomials one would encounter in high school math, i.e. over R. $\endgroup$ – Harry Stern Mar 4 '11 at 16:14

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