Polynomials are strange

Two-headed things

Polynomials are unexpectedly strange. They’re familiar from primary school, probably the first functions we study. But start studying abstract algebra or algebraic geometry, and they turn into two-headed beasts.

In abstract algebra you start to emphasize the algebraic structure of polynomials. You define addition of two polynomials much like you define component-wise addition of vectors. Multiplication gives polynomials a ring structure.

The textbook definition of the polynomial ring $ K[x]$ over a field $ K$ de-emphasizes that polynomials are functions. You could use vector notation to make this even clearer.

$ a x^2 + b x + c \equiv (a, b, c) \in K[x] $

Algebra books define the evaluation homomorphism to treat function-ness separately. Given the polynomial ring $ K[x]$ and $\alpha \in K$, the application:

$ ev_{\alpha}: K[x] \rightarrow K $

$ ev_{\alpha}(f) = f(\alpha) $

is easily proven to be a ring homomorphism.

You could do without the functional part of polynomials a lot of the time. But sometimes that part comes to the rescue in surprising ways.

Nullstellensatz

This famous theorem of Hilbert left me a bit confused initially.

Essentially, it says that the set of algebraic varieties in $ \mathbb{C}^n$ is in 1-1 correspondence with the set of radical ideals in $ \mathbb{C}[x_1,..,x_n]$.

An ideal of a ring is a subset of the ring that is closed under additions, and such that the product of an element in the ideal with any element in the ring, is also in the ideal.

A radical ideal is an ideal where, if some power of an element belongs, then the element belongs.

An example of a non-radical ideal is the ideal generated by $ f(x) = x^2$, aka $ (x^2)$. $ f(x) = x$ does not belong to the ideal, but $ f(x) = x^2$ does.

Fact: the ideal of polynomials that are zero on an algebraic variety $ V$ is a radical ideal.
This is almost trivial. Say $ f \in K[x]$, and $ f^k$ is zero over $ V$. Because the polynomial function takes values in a field, and a field is an integral domain, we must have $ f(x) = 0, \forall x \in V$.

There you have it, the function-ness saves the day. Otherwise, proving that an ideal is radical may be no picnic.

You noticed I said “the ideal of polynomials that are…”
There is generally more than one ideal whose polynomials are zero on a variety, but there is only one ideal that contains all the polynomials that are zero on the variety, and it happens to be radical.

Nullstellensatz gives us a 1-1 correspondence between the variety $ V$ and the radical ideal of polynomials that are zero on $ V$.

An exercise

This exercise from An Invitation to Algebraic Geometry by Karen E. Smith et. al. took me a while to solve:

Exercise 2.3.1 (abridged) Check that the ideal $ (xy, xz)$ is radical.

There’s an algebraic route to this, which seemed forbidding to me: to imagine some $ f^k$ in the ideal, and prove that $ f$ must also be in the ideal.

But the Nullstellensatz opens up a different path: Prove that $ (xy, xz)$ is the ideal of functions that are zero on the variety defined by $ (xy, xz)$.

This sounds silly to the point of tautology. One would expect that any ideal $ I$ should be the ideal for the variety it defines. Yet…

We can define a variety using a non-radical ideal. Think of the ideal $ I$ generated by $ (x^2 - 1)^2$. Its associated variety is $ V=\{1, -1\}$. But there is a radical ideal, $ I = (x^2 - 1)$, which is the ideal associated with the variety $ V$.

How do we prove that $ (xy, xz)$ is the ideal? We need to prove that any function that is zero on the variety must be in the ideal. The variety is easily seen to be $ V=\{(x,y,z) \in \mathbb{C}^3, x=0 \text{ or } y=z=0\}$

Say a polynomial $ g$ is zero on $ V$. Since $ x=0 \rightarrow g=0$, we can deduce that it must be divisible by $ x$. Indeed, by long division,

$ g = q x + r(y, z)$, and if we let $ x=0$, we see we must have $ r(y, z) = 0$.

Let’s study the quotient $ q$. Clearly, like $ g$, $ q=0 \text{ if } y=z=0$

Let’s try to divide $ q$ by $ y$. If it is fully divisible by $ y$, say $q = q_0 y$, we’re done, because $ g = x q_0y \rightarrow g \in (xy, xz)$.

Let’s assume there’s a remainder:

$ q = q_1y + r_1(x, z)$

Let’s divide again, this time $ r_1$ by $ z$.

$ q = q_1y + q_2z + r_2(x)$

Since $ q = 0$ when $ y=z=0$, we must have $ r_2(x) = 0$.

So,

$ q = q_1y + q_2z \rightarrow g = x q_1 y + x q_2z \rightarrow g \in (xy, xz)$

We conclude that $ (xy, xz)$ is the ideal for $ V$, so it’s radical.

Note that:

• The long division algorithm is algebraic, mostly
• But we used the function values to annul remainders

It still surprises me to switch back and forth on polynomials, more so than with other objects of dual nature.