Finite Algebras and Grobner Bases

Question

Background

Suppose that $A$ is a finite $\mathbb{R}$-algebra, that is, it is finite-dimensional as an vector space.
By a consequence of the Hilbert-basisatz, since $\mathbb{R}$ is Noetherian, then so is $A$, hence, in particular, there exists $f_1,\dots,f_n$ generating the ideal $\mathbb{R}[x]/(x^n)=\mathbb{R}(f_1,\dots,f_n)$. I assume these can be expressed using a Groebner basis.

Question

If $f_1,\dots,f_n$ is a Groebner basis of $A\triangleq \frac{\mathbb{R}}{(f_1,\dots,f_n)}$; $f_i \in \mathbb{R}$, and $N_1,\dots,_{N_n}>0$ is such that $$ Deg(f_1)={N_i}, $$ then does the set of all divisors of $f_1,\dots,f_n$ for a basis of $A$ as an $\mathbb{R}$-module?

Example

In the special case that $A=\mathbb{R}[x]/(x^n)\cong \mathbb{R}^{n}$ as an vector space over $\mathbb{R}$. This is indeed true and $$ \mathcal{B}= \{x^i \}_{i=0}^{N_n}\qquad ;N_n=n , $$ forms a basis of $A$ as a $\mathbb{R}$-algebra.

Answer 1

In your "Question" section, I don't know what you mean by "if $f_1,\ldots, f_n$ is a Grobner basis of $A$". A Grobner basis is usually defined for an ideal of a polynomial ring over a field, not just any algebra.

Maybe you mean that $A$ is a quotient of a polynomial ring $\mathbb R[x_1,\ldots, x_m]/I = A$, as in your example, and you mean that $\{f_i\}$ is a Grobner basis for the ideal $I = (f_1,\ldots, f_n)$. Then no, $\{f_i\}$ is not a basis for $A$, it's the opposite of a basis - all of its elements are 0 in $A$.

You have mentioned $\mathbb R[x]/(x^n)$ and said it is "indeed true" that $\{x^i\}_{i=0}^{n-1}$ forms a basis for this algebra. But was it a Grobner basis in the first place? Why do you think this/what does it mean?

ADDED:

OK I think I understand what you're trying to ask. Maybe your questions is:

How can a Grobner basis for an ideal $I$ help us find a basis for $A = \mathbb R[x_1,\ldots, x_n]/I$? (Possibly in the special case that $A$ is finite dimensional.)

I'll answer this by suggesting a couple examples.

1. In one variable, $\mathbb R[x]$ is a PID so all ideals are of the form $(f)$. This is a rather trivial use of Grobner bases, but the pattern that extends to multiple variables is that a basis for the quotient $A$ is given by power of $x$ less than the degree of $f$.

2. An example in two variables is the ideal $I = (x^2y^2 + xy + 1, x^3, y^4)$. One Grobner basis for this is $\{x^2y^2, x^3, y^4\}$. What is a basis for the quotient? Any monomial $x^iy^j$ for a pair $(i,j)$ such that $i<3$ and $j<4$ and if $i\geq 2$ then $j < 2$. In other words $\{1,y,x,x^2,xy,y^2,x^2y,xy^2,y^3,xy^3\}$.

As you can see, the conditions that come up in 2 are not about divisibility, but just about bounds on what degree each variable can have. When you have mixed terms of more variables the constraints become more complicated, which is where the monomial orders come in.

Lastly I'll mention that it's not important that $A$ be finite dimensional. You can still find a basis of the quotient using a grobner basis for the ideal. You could do this in $A$ for instance when you remove the assumption that $y^4 = 0$. Then a basis for $A$ will be $y^i, xy^i, x^2, x^2y$ where $i \geq 0$ is any power of $y$.

Answer 2

I didn't completely understand your question, as there are many obscure things as Ben pointed out.

However, I assume that you are asking something like this: I have a commutative algebra $A$ over $\mathbb{R}$ which is finite-dimensional as an $\mathbb{R}$-vector space. In particular, it has to be of the form $A\cong\mathbb{R}[X_1,\ldots,X_n]/(f_1,\ldots,f_m)$ for some polynomials $f_1,\ldots,f_m$. I fix a monomial order and I assume that $\{f_1,\ldots,f_m\}$ is a Groebner basis for the ideal they generate. Is it true that the set $\mathcal{B}$ containing the equivalence classes of all the divisors of the polynomials $f_1,\ldots,f_m$ forms a basis for $A$ as a vector space over $\mathbb{R}$?

The first trivial answer is of course, not: the classes of $f_1,\ldots,f_m$ are all $0$ and they are in $\mathcal{B}$. So, let us assume you want only the proper divisors (whence excluding $f_1,\ldots,f_m$ and $1$). The answer is still no: $\{x^2+1\}$ is a Groebner basis for $(x^2+1)$ in $\mathbb{R}[x]$ but it has no divisors at all.

Notice that you have to exclude $1$, otherwise $\{x^2-1\}$ is a Groebner basis for $(x^2-1)$ in $\mathbb{R}[x]$ but $\mathcal{B}=\{[x-1],[x+1],[1]\}$ which is not free over $\mathbb{R}$ (with $[\;\cdot\;]$ I denote the equivalence class in the quotient).

Edit: It doesn't even work over an algebraically closed field: $\{X^3-1\}$ is a Groebner basis for $(X^3-1)$ in $\mathbb{C}[X]$, however the elements in $$\mathcal{B}=\{[X-1],[X-\xi],[X-\xi^2],[X^2+X+1],[X^2 + \xi^2 X+\xi],[X^2+\xi X+\xi^2]\}$$ are not linearly independent, where $\xi$ is a root of $X^2+X+1$. Indeed $$[X-\xi]+[X-\xi^2]-\frac{1}{3}\Big(2[X^2+X+1]-[X^2+\xi X+\xi^2]-[X^2+\xi^2X+\xi]\Big) = [X] = \frac{1}{3}\Big([X-1]+[X-\xi]+[X-\xi^2]\Big)$$ which are two expressions of $[X]$ as linear combination of elements of $\mathcal{B}$.