Hilbert's nullstellensatz'^H'*15 basis theorem

by math_explorer, Mar 26, 2015, 5:34 AM

I was going to make a long weird walkthrough of Hilbert's Nullstellensatz, but it looks like it's too long for AoPS, so I'm going to extract one of the parts. In other words I'm irresponsible and don't remember how to prove prerequisites from a long time ago:

Hilbert basis theorem. If a commutative ring $R$ is Noetherian, then the polynomial ring $R[x]$ is Noetherian.

Terminology dump in case you forget like I do, or if you would like to follow along without knowing anything about abstract algebra, although I have to admit this is not very likely:

A commutative ring is a thing in which you can add, subtract, and multiply; addition and multiplication are both associative and commutative; there's a 0 and a 1 (additive and multiplicative identities, respectively); and multiplication distributes.
Rings in general don't have to have commutative multiplication, and you can actually define left- and right-Noetherian noncommutative rings, but for now we'll just deal with the intuitive commutative case. I was going to write that I would use "ring" just to mean commutative ring like lots of textbooks do, but I realized I don't actually refer to the word much at all, and I should hold myself to higher notational standards than math textbooks. (Fulton does this four lines into Chapter 1:)

Quote:

When we speak of a ring, we shall always mean a commutative ring with a multiplicative identity.

(Jacobson is even sneakier, putting this in the middle of a chapter — )

Jacobson Chapter 2 Section 10 wrote:

For the remainder of this chapter—except in section 2.17 and in an occaional exercise—all rings will be commutative and the word "ring" will be synonymous with "commutative ring."

(One of those magical dependencies you say once and only once that really annoy me. Blink and you'll miss it. Don't code like this at home.)
The polynomial ring $R[x]$ is the ring of polynomials with coefficients in $R$ . You treat each polynomial as its own algebraic thing, and add, subtract, and multiply them as polynomials. I don't know what I should explain here, oops.
A commutative ring is Noetherian if there is no infinite chain of ascending ideals...
Oh wait I need to explain an ideal. An ideal $I$ of a commutative ring $R$ is a subset of $R$ that is closed under addition (with another element of $I$ ) and under multiplication by any element of $R$ .
The last bit is important. For example, if we have the polynomial ring $\mathbb{Q}[x]$ then the set $S$ of all polynomials with only even-degree terms is closed under addition and under multiplication by itself, but not by any element of $R$ ; for example $x^2 \in S$ and $x \in R$ but $x^2 \cdot x \not\in S$ . So $S$ is not an ideal.
Instead, the set of all polynomials without terms with degree 1 or 0 is an ideal, the smallest one that contains $x^2$ .
In practice, ideals are intuitively something like "linear combinations of a bunch of stuff." For example, in a double polynomial ring $R[X_1][X_2]$ , all elements of the form $a(X_1^2 + 42) + b(X_2 - 1337)$ would form an ideal.
Back to defining Noetherian... "Ascending ideals" are just ideals that get bigger, $I_1 \subseteq I_2 \subseteq \cdots$ . Rigorously, a commutative ring is Noetherian if any infinite chain of ideals $I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots$ is eventually constant.
So "no-infinite-chain-of-ascending-ideals" is not technically right; it's a more concise and intuitive (to me) way to almost express the alternate idea that "no infinite strictly ascending chain of ideals $I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \cdots$ exists." But it turns out "ascending-chains-of-ideals-are-eventually-constant" is slightly easier to work with.

Both Wikipedia and Dummit-Foote use the alternate characterization of Noetherian rings, that every ideal contained in one is finitely generated as a module, to prove the Hilbert basis theorem. I'm trying to not make that detour because the Hilbert's Nullstellensatz walkthrough this was extracted from only needs this condition on ascending-chains, so I'm writing this proof on my own. The upside is I get some abstract experience and I can claim that this post is at least somewhat original. The downside is that it may be buggy.

Here goes!

Proof: Let $I_1 \subseteq I_2 \subseteq \cdots$ be an infinite chain of ascending ideals in $R[x]$ . We want to prove it's eventually constant.

Define sets $I_{i,j}$ as the set of all leading coefficients of polynomials in $I_i$ with degree exactly $j$ , where we suppose for now that the zero polynomial has degree whatever we want it to be and therefore include $0$ in every $I_{i,j}$ .

We may verify some facts.

Each $I_{i,j}$ is an ideal in $R$ . Just add the corresponding polynomials or multiply them by a constant in $R[x]$ to get the properties an ideal must satisfy.
For all $i, j$ , we have $I_{i,j} \subseteq I_{i,j+1}$ . Just multiply each polynomial corresponding to an element $e$ of $I_{i,j}$ by $x$ in $R[x]$ to see that $e \in I_{i,j+1}$ .
For all $i, j$ , we have $I_{i,j} \subseteq I_{i+1,j}$ . True since the set of degree- $j$ polynomials in $I_i$ is a subset of the set of degree- $j$ polynomials in $I_{i+1}$ .

Since $R$ is Noetherian, for any $i$ , the chain $I_{i,0} \subseteq I_{i,1} \subseteq I_{i,2} \subseteq \cdots$ is eventually constant. Let's abuse notation and denote as $I_{i,\infty}$ the ideal that the chain is eventually constantly equal to.

Since $I_{i,j}$ is contained inside $I_{i,\infty}$ for every $j$ and $I_{i,\infty}$ contains no elements outside all $I_{i,j}$ , we know that $I_{i,\infty}$ is the ideal of all leading coefficients of polynomials in $I_i$ , an ideal which is actually also considered by Wikipedia and Dummit-Foote in their proofs. This definition makes it easy to see that $I_{i,\infty} \subseteq I_{i+1,\infty}$ for all $i$ .

Then $I_{1,\infty} \subseteq I_{2,\infty} \subseteq I_{3,\infty} \subseteq \cdots$ is also an ascending chain of ideals, so it is eventually constant.

Suppose $I_{i,\infty}$ is constant when $i \geq m$ . Pick $n$ such that $I_{m,\infty} = I_{m,n}$ . Then for all $i \geq m, j \geq n$ , $I_{m,n} \subseteq I_{i,n} \subseteq I_{i,j} \subseteq I_{i,\infty} = I_{m,\infty}$ shows that $I_{m,n} = I_{i,j}$ . In particular, since $i, j$ were arbitrary, $I_{i,j} = I_{i+1,j}$ for all $i \geq m, j \geq n$ .

Also, for each $0 \leq j \leq n$ , the chain $I_{1,j} \subseteq I_{2,j} \subseteq I_{3,j} \subseteq \cdots$ is eventually constant as well. Suppose each such chain is constant after some $I_{m_j, j}$ . We are only considering finitely many $j$ here, so we may let $M$ be the max of all these $m_j$ and $m$ .

So when $i \geq M$ , we have $I_{i,j} = I_{i+1,j}$ for all $j$ . We claim that this implies $I_i = I_{i+1}$ , so $I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots$ is constant after $I_M$ , which finishes our proof.

We argue by contradiction: if $I_i \subsetneq I_{i+1}$ , take a polynomial $P \in I_{i+1} \setminus I_i$ with minimal degree and suppose its leading term is $cx^d$ .

This implies that $c \in I_{i+1,d}$ . But because $I_{i,d} = I_{i+1,d}$ , there was already a polynomial $Q$ with leading term $cx^d$ in $I_i$ , and $Q$ is distinct from $P$ . Then $Q \in I_{i+1}$ too, and $P - Q \in I_{i+1}$ , but since $P \not\in I_i$ , we know $P - Q \not\in I_i$ . Therefore $P - Q \in I_{i+1} \setminus I_i$ , and yet the leading terms of $P$ and $Q$ cancel out, which means $P - Q$ has degree lower than $Q$ , contradicting the minimality of the degree of $P$ .

Therefore $I_{i+1} = I_i$ for all $i \geq M$ , proving that $I_i$ is eventually constant and that $R[x]$ is Noetherian.

-- end proof --

The obvious corollary should also be mentioned for the sake of completeness...

If a commutative ring $R$ is Noetherian, then the polynomial ring $R[x_1, x_2, \ldots, x_n]$ is Noetherian.

Proof. Induct on $n$ . If $R[x_1, \ldots, x_n]$ is Noetherian then $R[x_1, \ldots, x_n][x_{n+1}] = R[x_1, \ldots, x_{n+1}]$ is Noetherian.

abstract algebra

algebra

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Hilbert's nullstellensatz'^H'*15 basis theorem

by math_explorer, Mar 26, 2015, 5:34 AM

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