# How Do You Apply Noether Normalization to a Polynomial Ring Ideal?

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• gurilupi
In summary, In this conversation, the discussion revolved around finding a Noether normalization for an ideal generated by maximal minors of a ##2 \times 3## matrix. The proof involves using an induction and a substitution method to transform the ideal into a polynomial relation. The details of the proof can be found in the sources mentioned and it is recommended to work through the example step by step for better understanding.
gurilupi
Suppose ##I \subseteq k[X_{1}, X_{2}, X_{3}, X_{4}]## be the ideal generated by the maximal minors of the ##2 \times 3## matrix
$$\begin{pmatrix} X_1 & X_2 & X_3\\ X_2 & X_3 & X_4 \end{pmatrix}.$$

I have to find a Noether normalization ##k[Y_1, Y_2, Y_3, Y_4] \subseteq k[X_1, X_2, X_3, X_4]## with ##I \cap k[Y_1, Y_2, Y_3, Y_4] = (Y_1, \ldots, Y_r)## for a suitable ##r##.

I've done: The maximal minors are the determinant(s) of the largest submatrices, i.e. in this case all ##2 \times 2## submatrices which then are (by deleting a column): ##\begin{pmatrix}
X_1 & X_2\\
X_2 & X_3
\end{pmatrix}, \begin{pmatrix}
X_1 & X_3\\
X_2 & X_4
\end{pmatrix}, \begin{pmatrix}
X_2 & X_3\\
X_3 & X_4
\end{pmatrix}.## Then, taking determinants we get ##I = (X_1 X_3 - X_{2}^{2}, X_1 X_4 - X_3 X_2, X_2 X_4 - X_{3}^{2})##.

The next step would be to use the constructive proof of Noether's Normalization Lemma. However, I can't seem to understand the entire procedure of that proof and how to apply it to this problem. Perhaps if someone can illustrate this process, then I will better understand it after seeing it done.

The proof I have uses an induction. The following lemma seems crucial:

Let ##F\in k[X_1,\ldots,X_n]-\{0\}##. Then there is a substitution ##X_i=Y_i+X_n^{r_i}## for ##i=1,\ldots,n-1## and suited ##r_i## such that
$$F=aX_n^m+\rho_1 X_n^{m-1}+\ldots+\rho_m \quad (a\in k-\{0\},\rho_i\in k[Y_1,\ldots,Y_{n-1}])$$
The ##Y_i## are all inductively constructed according to this lemma.

The best approach is probably to run through the proof step by step for the given example.

Thanks for the answer. The proof of Nother's Normalization Lemma that I have also uses induction. The issue is that the hypothesis assumes a finitely generated algebra ##A##. So what is my finitely generated algebra and what are the generators?

gurilupi said:
The proof of Nother's Normalization Lemma that I have also uses induction.
Kunz? It's difficult to talk about a technical proof with two different sources.

fresh_42 said:
Kunz? It's difficult to talk about a technical proof with two different sources.

No, it's from Atiyah & Macdonald Introduction to commutative algebra.

Damn, that's an exercise ...

My proof deals with infinite fields as special case, i.e. I have what Atiyah calls the different proof. Makes communication not easier.

That's not an issue since in the course we are assuming our field to be algebraically closed, i.e. infinite. Thus for the purpose of the problem statement we do have that ##k## is infinite. The issue that I have here is that I never seen the actual process of Noether normalization in practice. If the details are not too lengthy, perhaps you can write them out? I suspect that the computations are quite algorithmic and once one sees the process, then it becomes quite easy to replicate.

Edit: BTW, Atiyah also assume ##k## infinite. For finite ##k## one needs a different proof.

gurilupi said:
Thanks for the answer. The proof of Nother's Normalization Lemma that I have also uses induction. The issue is that the hypothesis assumes a finitely generated algebra ##A##. So what is my finitely generated algebra and what are the generators?

Your ##k##-algebra is your ideal ##I##. You have listed its generators. Do you see a polynomial relation between the generators?

gurilupi
Infrared said:
Your ##k##-algebra is your ideal ##I##. You have listed its generators. Do you see a polynomial relation between the generators?

No, I don't see a polynomial relation between the generators. I suppose you mean that we can somehow remove some indeterminants by substitution?

gurilupi said:
No, I don't see a polynomial relation between the generators. I suppose you mean that we can somehow remove some indeterminants by substitution?
##(Y+X_1)\cdot\ldots\cdot(Y+X_n) =\sum_{k=0}^n \sigma_k(\mathbf{X})Y^{n-k} ## with ##\sigma_k(\mathbf{X})=\sigma_k(X_1,\ldots,X_n)=
\sum_{1\leq j_1<\ldots <j_k\leq n}X_{j_1}X_{j_2}\ldots X_{j_k}## could help.

Thanks for the link but I am uncertain where to even start the normalization process. The proof you linked says: induct on the number of generators of the ##k##-algebra. So in our case on the three generators of the ideal. Then, you pick one generator, say the first and check whether ##I = k[X_1 X_3 -X_{2}^{2}]## and whether the generator is transcendental. I suppose it’s algebraic and thus must have a minimal polynomial, say ##f##. Then, ##I \cong k[X_1, X_2, X_3, X_4] / (f)##.

And this is where my understanding breaks down; the next step assumes ##I = k[c_1, c_2, c_3, c_4]##, where the ##c_i## are the generators. How is the ideal the polynomial ring in the four variables ##c_i##?

In post #1, I don't even understand your second sentence. What is the definition of "a noether normalization"? And I would think your k algebra is the quotient of k[x1,..,x4] by your ideal I. (I itself is not a k algebra, in the usual sense, since it does not contain k.) But when I even talk about a noether normalization, I am always given a k algebra A to begin with. The fact that in post #3 you yourself ask what the k algebra is, suggests to me that you do not understand what you are being asked to do, which of course you admit. But could you just quote the problem you are given for us word for word?

Noether's normalization theorem is the process of decomposing a finitely generated algebra extension of k into two stages. one which is purely transcendental, followed by one which is integral. You cannot begin until you determine what the k algebra is that is given. This must be part of the data given in the problem.Presumably the f.g. k algebra is the quotient algebra A = k[X]/I. Then you want to find a sub algebra B which is isomorphic to a polynomial algebra, and such that A is integral over B.

Now one way to do this would be to find a sub algebra C of k[X], which is also a polynomial algebra in 4 variables Yj over k, and such that two things are true:

1) I intersect C = J is the ideal of C generated by some of the variables Yj;
so that the quotient B = C/J is again a polynomial k algebra;

2) and such that the extension A over B is module finite, hence integral.

I.e. your original given f.g. k algebra is presumably A = k[X]/I, and your sub algebra is B = k[Y]/J = C/J. By definition of J, the injection C-->k[X], induces an injection of B = C/J -->A.

Then A is finite over B and B is a polynomial algebra. So by the way your problem is set up, "all" you need is a polynomial sub algebra C = k[Y1,Y2,Y3,Y4] of k[X1,...,X4], with the Y's algebraically independent, and meeting I in an ideal J generated by some of the Yj, and such that A is finite over B = C/J.

I doubt if the general proof of the normalization theorem need be applied, since this is a concrete example, but it might help. Forgive me, as I have not really thought about your problem. (Does this belong in the homework section?) Were you told as a hint to use the "constructive proof of noether normalization"? were you given a reference for that "constructive proof"?

but anyway, if this interpretation is correct, then of course X1,...,X4 are your original generators for k[X] hence also for A = k[X]/I.

Last edited:
Infrared

## 1. What is Noether Normalization?

Noether Normalization is a mathematical method used in commutative algebra to simplify polynomial rings. It was developed by German mathematician Emmy Noether in the early 20th century and has since become an important tool in algebraic geometry and related fields.

## 2. How does Noether Normalization work?

Noether Normalization works by finding a finite set of algebraically independent elements in a given polynomial ring, known as the Noether Normalization basis. This basis can then be used to transform the original polynomial ring into a simpler form, making it easier to study and analyze.

## 3. What are the applications of Noether Normalization?

Noether Normalization has various applications in mathematics, including algebraic geometry, commutative algebra, and number theory. It is used to study the structure of algebraic varieties, solve systems of polynomial equations, and prove important theorems in these fields.

## 4. What are the limitations of Noether Normalization?

Noether Normalization is not applicable to all polynomial rings. It only works for polynomial rings that satisfy certain conditions, such as being finitely generated and having a finite transcendence degree. Additionally, the process of finding the Noether Normalization basis can be computationally intensive for large polynomial rings.

## 5. Are there any variations of Noether Normalization?

Yes, there are variations of Noether Normalization that have been developed over the years. These include the Macaulay Normalization, Zariski Normalization, and the Nagata Normalization. Each variation has its own specific conditions and applications, but they all share the same goal of simplifying polynomial rings.

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