# A Problem Regarding Polynomials

Let $$F$$ be a field, unique $$a_0, a_1, ... ,a_n \in F$$, and let $$b_0, b_1, ... ,b_n$$ be some elements in $$F$$ (not necessarily different).

A) Prove there exists a unique $$f \in\mathbb{F}[x]$$, with $$deg f \le n$$, and $$f(a_i) = b_i$$ for all $$0\le i\le n$$.

B) Let $$F = \mathbb C$$, and let $$b_i = \overline{a_i}$$. Prove there exists $$f \in\mathbb{R}[x]$$ of some degree, such that $$f(a_i) = b_i$$ for all $$i$$.

Hint: consider the set $$A = {{a_i}}\cup{{\overline{a_i}}}$$, and by using A), prove that $$f = \overline{f}$$.

I've proven A), But not B). I tried to define a linear transformation, but since $$a_i$$ might be equal to $$\overline{a_i}$$, I'm unsure of what N-tuple I'd pick. Am I looking at this from the wrong direction?

• There is a problem in the hint, as what you would want to prove is that $f=\tau\circ f\circ\tau$ where $\tau(z):=\overline{z}$. The identity $f=\overline{f}(=\tau\circ f)$ for a polynomial $f\in\mathbb{C}[x]$ implies $f=0$. And I don't see why it would be a problem if $a_i=\overline{a_i}$, in this case you assign to it the value $b_i=\overline{b_i}$ (part A works for any $a_1,\ldots,a_n$, not necessarily distinct, if you add the extra condition that $b_i=b_j$ whenever $a_i=a_j$). Commented Mar 17, 2023 at 18:03
• @imtrying46 are you sure you're thinking of elements in the polynomial ring $\mathbb C[x]$? My guess is you are thinking of polynomial functions from complex analysis, in which case the implication isn't $f=0$ per se but constant $f$ (and evidently real). The mention of $f=\tau\circ f\circ\tau$ also looks like complex analysis; the notion of function composition $f\circ \tau$ doesn't register since $f$ is not a function but an element in $\mathbb C[x]$. Commented Mar 17, 2023 at 22:39
• Yeah. The problem I'm having is that f should be in the Reals, I'm unsure how to approach that.
– FNB
Commented Mar 18, 2023 at 7:22
• @user8675309 You’re right, of course if $f=\overline{f}$ then $f$ is constant real, not necessarily $0$. But I don’t get your problem about distinguishing polynomials from say holomorphic functions. If $R$ denotes the ring of holomorphic functions, then there is an injective ring homomorphism $\mathbb{C}[x]\to R$, assigning to a polynomial the associated function. So it makes perfect sense to regard polynomials as holomorphic functions, and use complex analysis on them. For example, the fact that every polynomial in $\mathbb{C}[x]$ has a root is typically proven by using complex analysis. Commented Mar 19, 2023 at 9:54
• @imtrying46 The issue is that $f\in \mathbb C[x]$ which is a field with a single element "$x$" adjoined to it hence a polynomial ring; "x" is compatible w/ linear combinations w/ itself and elements in $\mathbb C$ but there is no additional element $\bar x$. You can do a homomorphism $\phi$ into the ring of entire functions if you like but $\phi\circ f\neq f$. Yet $\bar f \in \mathbb C[x]$ which is to say $\bar f$ is $f$ with all scalars $c_k \in \mathbb C$ conjugated, so the official hint is fine. Put differently: you are working in a different ring than the OP which causes discrepancies. Commented Mar 19, 2023 at 16:56

An explicit way to do (A) and (B) is via an Vandermonde matrix. For (B), let $$S:= \big\{a_1,\dots, a_n\big\}\cup\big\{\overline a_1,\dots, \overline a_n\big\}$$ with $$r:=\big \vert S\big \vert$$ and consider the degree $$r$$ polynomial $$c_0 + c_1 x + \dots c_r x^r = p(z) \in \mathbb C[x]$$.
$$V\mathbf c = \mathbf b$$ where the $$r\times r$$ Vandermonde matrix has the moment curve $$[1, s, s^2, ..., s^{r-1}]$$ for $$s\in S$$ on each row, each of which is unique, each row of $$\mathbf b$$ contains the value of $$p$$ when evaluated at $$s$$ (technically a substitution homomorphism $$\phi:\mathbb C[x]\rightarrow \mathbb C$$) and $$\mathbf c$$ has the coefficients of $$p$$ to be solved for. $$V$$ is invertible since all rows are unique hence $$\mathbf c$$ is uniquely specified. But for for some permutation matrix $$P$$
$$\big(P V\big)\overline {\mathbf c} =\overline V\overline {\mathbf c} = \overline{\mathbf b}=P\mathbf b\implies V\overline {\mathbf c}=\mathbf b$$
$$\implies \big(\overline {\mathbf c}-\mathbf c\big)\in \ker V\implies \mathbf c\in \mathbb R^r$$, i.e. the coefficients of $$p$$ are real since $$V$$ is injective
Let $$\# A=2n-m.$$ That means $$a_k=\overline{a_k}$$ for $$m$$ values of $$k.$$ By (A) there is a unique polynomial $$p(z)$$ of degree $$2n-m-1$$ such that $$p(a_i)=\overline{a_i}$$ and $$p(\overline{a_i})=a_i.$$ Consider the polynomial $$q(z)$$ with the coefficients that are complex conjugate of the coefficients of $$p(z),$$ i.e. $$q(z)=\overline{p(\overline{z})}.$$ Then $$q(a_i)=\overline{a_i}$$ and $$q(\overline{a_i})=a_i$$ for any $$i.$$ By the unicity we get $$q(z)=p(z).$$ Hence the coefficients of $$p(z)$$ are real numbers.