# When is the zero polynomial the only zero function?

Like all mathematics, the first thing to do is set the landsape in which you wish to work in. Let $K$ be a field and let $K[x_1,...,x_n]$ be the commutative polynomial ring with $n$ variables (if you are familiar with this ring I would skip the next paragraph)

What do the elements of  $K[x_1,...,x_n]$ look like? Taking $n =3$ it is not hard to come up with elements, for example $x_1x_3 +{x_2}^4 + {x_2}^3{x_3}^2$. One can quite easily start conjuring up polynomials for a given $n$ but a more interesting approach to answer the question, can we find a ‘generating’ set for our polynomials? This brings one to the notion of a moninomial which is simply a product of the form $x_1^{\alpha_1} ... x_n^{\alpha_n}$. We can simplify the notation by starting with an $n$-tuple $\alpha = (\alpha_1,..., \alpha_n)$ of non-negative integers and letting $X^\alpha = x_1^{\alpha_1} ... x_n^{\alpha_n}$ . Clearly a poynomial $f \in K[x_1,...,x_n]$ can be written as a finite linear combination of moninomials with the general form $f = \sum_{\alpha \in A} C_\alpha X^\alpha$ where $C_\alpha \in K$ and $A \subset {\mathbb{N}_{0}}^n$ with $A$ also being finite. Without knowing what a polynomial is aprior this is the method used to define what a polynomial is with $n$ variables.

$K[x_1,...,x_n]$ is a ring with addition and multiplicative defined how you would expect and as $K$ is a field the ring is commutative. Given $f \in K[x_1,...,x_n]$ we can think of it as an algebraic element of our ring or a function $f : K^n \longrightarrow K$ by swapping $x_i$ with some $a_i \in K$ for all $i$. This gives us two notions of zero; the zero element of $K[x_1,...,x_n]$ and a zero function from $K^n \longrightarrow K$.

To emphasise the difference let us take $K = \mathbb{F}_2$, $n =1$ and look at the polynomials $f = 0$ and $g = x(x-1)$ in $K[x]$. As algebraic elements clearly $f$ is the zero element and $g$ isn’t but as functions $g(0) = 0$ and $g(1) = 0$ so we can conclude both $f$ and $g$ are zero functions. To wrap up we have two distinct elements that are both zero functions $K^n \longrightarrow K$.

This invokes a big question and one in which we will answer: When is the zero polynomial the only zero function? (I didn’t lead you with false pretenses by the title). The answer depends on the cardinality of $K$.

If $K$ is finite of size $q$ by Lagrange’s Theorem and looking at the multiplicative group $K\backslash \{0\}$ one can show $x^q = x$ for all $x \in K$ so the polynomial $f = x^q- x$ is a zero function but is distinct from the zero element in $K[x_1,...,x_n]$. If $n > 1$ we can define $f = x_1^q - x_1$. This tells us that when $K$ is finite we can always come up with elements in $K[x_1,...,x_n]$ that are non-zero and are a zero function $K^n \longrightarrow K$ so there is no unique zero function.

What about the case when $K$ is infinite? Well…

Theorem: Let $K$ be an infinite field, let $f \in K[x_1,...,x_n]$. Then, $f = 0$ if and only if $f : K^n \longrightarrow K$ is the zero function.

Proof of Theorem: Let us begin with the ‘easy’ direction, suppose $f = 0 \in K[x_1,...,x_n]$ trivially we must have $f(\alpha) = 0$ for all $\alpha \in K^n$. The other direction requires us to show that if for some $f \in K[x_1,...,x_n]$ such that $f(\alpha) = 0$ for all $\alpha \in K^n$ then $f$ is the zero element in $K[x_1,...,x_n]$.

We will use induction on the number of variabels $n$. When $n = 1$ a non-zero polynomial in $K[x]$ of degree $m$ has at most $m$ roots. Suppose $f \in K[x]$ such that $f(\alpha) = 0$ for all $\alpha \in K$ and since $K$ is infinite $f$ must have infintely many roots and as the degree of $f$ must be finite we must conclude that $f = 0$.

Now assuming the inductive hypothesis for $n - 1$ let $f \in K[x_1,...,x_n]$ be a polynomial such that $f(\alpha) = 0$ for all $\alpha \in K^n$. By collect the various powers of $x_n$ we can write $f$ in the form of $f = \sum_{i}^{N} g_i(x_1,...,x_{n-1})x_n^i$ where $N$ is the highest power of $x_n$ that appears in $f$ and $g_i \in K[x_1,...,x_{n-1} ]$. If we show that $g_i$ is the zero polynomial in $n - 1$ variables this will force $f = 0$ in $K[x_1,...,x_n]$.

If we fix $(\alpha_1,...,\alpha_{n-1}) \in K^{n-1}$ we get a polynomial $f(\alpha_1,...,\alpha_{n-1}, x_n) \in K[x_n]$ and by our hypothesis this polynomial vanishes for all $\alpha_n \in K$. Using our above representation of $f$ we see the coefficients of $f(\alpha_1,...,\alpha_{n-1}, x_n)$ are $g_i ( x_1,...,x_{n-1} )$ and hence $g_i(\alpha_1,...,\alpha_{n-1}) = 0$ for all $i$. As $(\alpha_1,...,\alpha_{n-1}) \in K^{n-1}$ is chosen arbitraty $g_i$ is the zero function $K^{n-1} \longrightarrow K$ and so by the inductive hypothesis $g_i$ is the zero polynomial in $K[x_1,..., x_n]$.

Bringing in both of these results lets us conclude that if a polynomial zero function $K^n \longrightarrow K$ is the zero element of $K[x_1,...,x_n]$ then we must have that the field $K$ is infinite. I find this quite a remarkable result in the fact that knowledge about a zero function gives you an indication on the cardinality of $K$.

The theorem we have just proved also lets us conclude that when $K$ is infinite polynomials are unique functions or more precisely for $f ,g \in K[x_1,...,x_n]$ $f = g$ in $K[x_1,...,x_n]$ if and only if $f: K^n \longrightarrow K$ and $g : K^n \longrightarrow K$ are the same functions. I leave the proof of this for the reader and as a hint consider the polynomial $f - g$.