Are two polynomials which take the same values when evaluated equal?

Question

Two polynomials are considered equal if they have equal coefficients of corresponding powers of the independent variable, after like terms are combined. If two polynomials are equal in this sense, then they are equal as functions; i.e., they give equal results for equal values of the independent variable. If, conversely, two polynomial functions differ in their corresponding coefficients, do they always have different values for the same independent variable? Is this true at least for the infinite field of integers, and where I can find a proof of that?

Answer 1

This is true for infinite integral domains.

The proof is very simple: over an infinite integral domain, a polynomial function has a finite number of roots, bounded by its degree.

Now, if two polynomials $f$ and $g$ have the same value for an infinite number of values of the variable, the polynomial $f-g$ has an infinite number of roots, which cannot happen since $\deg(f-g)\le\max(\deg f,\deg g)$.

However, for finite fields, this is false. For instance, over the field $\mathbf F_p$, the polynomial $X^p-X\ne 0$, but the polynomial function $x\mapsto x^p-x$ is the same as the $0$ function.

Answer 2

It is true for polynomials $f(x)$ over any infinite integral domain $R$. It is enough to show that if a polynomial $f(x)$ is identically zero, all its coefficients are zero.Let $n$ be the degree of the polynomial, then the polynomial has $n+1$ coefficients. Take $n+1$ distinct elements $a_0,...,a_n$ in the ring and plug it into $f(x)$. You get $n+1$ homogeneous linear equations whose unknowns are the coefficients of $f$. The determinant of that system is the Vandermond determinant which is equal to the product of all $a_i-a_j$, $i>j$. Since $R$ is a domain, the determinant is not zero, and the system has unique solution $(0,0,...)$.So for a given two polynomials $f, g$ of degrees $m\le n$ to prove that the corresponding functions are different you need that $R$ has at least $n+1$ different elements.

Answer 3

The other answers already showed you that there are different polynomials with the same polynomial function in finite fields. I'd like to shed some light on why this is sensible. After all, why define polynomials if not to study polynomial functions, so why should they have different properties?

The thing is, a polynmial $f\in R[X]$ does not only define a polynomial function $f:R\to R$. Way better, if $S$ is a ring with $R\subseteq S$, then $f$ even defines a function $f:S\to S$. Essentially, we can not only evaluate a function at elements of the underlying ring. We can also evaluate them at elements of any larger ring in which $R$ is contained as well. For instance, the ring of $n\times n$ matrices over $\mathbb R$ contains $\mathbb R$ if we identify $r\hat=\operatorname{diag}(r,\dots,r)$. And in this way, we can insert matrices into any real polynomial, not just real numbers. This is the reason why we are okay with polynomials over finite fields which have different coefficients while taking on the same values in the entire field: They may not take on the same values everywhere in a larger ring.

For instance, take the polynomial $f=X^2+X\in\mathbb F_2[X]$. It is zero everwhyere in $\mathbb F_2$ even though it is not the zero-polynomial. But if we insert the matrix

$$A=\begin{pmatrix}0&1\\1&0\end{pmatrix},$$

we get

$$f(A)=A^2+A=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&1\\1&1\end{pmatrix}.$$

So if we insert an element of a larger ring, in this case the ring of $2\times2$ matrices over $\mathbb F_2$, the associated polynomial is no longer identical to the polynomial function of the zero-polynomial. This is why it makes sense to allow different polynomials which give the same output everywhere in the underlying ring. They might not give the same output everywhere in a larger ring.