2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
7
$\begingroup$

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 $\:\pm 1$.

$\endgroup$
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ @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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.