# Equality of polynomials: formal vs. functional

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.

• Yes. Consider g(x) = A - B. – Harry Stern Mar 4 '11 at 15:11
• You just said A=B for all x. x is the only variable, so yes. – TROLLHUNTER Mar 4 '11 at 15:13
• As Anjan mentions below, it depends on what kind of things the $a_k$ and $b_k$ are. Would you mind clarifying? – Jason DeVito Mar 4 '11 at 15:21
• @Jason DeVito: $a_k$ and $b_k$ are just coefficients. They're independent of $x$. – FUZxxl Mar 4 '11 at 15:37
• 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? – 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 ${\mathbb{Z}/2\mathbb{Z}}[X]$.
• @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 4 '11 at 15:25