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.



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