# Finitely generated algebra finding generators and quotients

The following easy lemma is known to me: Let $$R$$ be a commutative ring and $$A$$ be a finitely generated $$R$$-algebra. Then it is isomorphic to a quotient of a polynomial ring $$R[x_1,\ldots,x_n]$$ of finitely many variables.

There is something in ring theory I never understood and it seems to come back in algebraic geometry right now while looking at irreducible components of affine varieties.

For example I have the following rings:

$$A=\{f\in \Bbb Q[X]\mid f'(0)=0 \}$$

and

$$B=\{f\in \Bbb Q[X]\mid f(0)=f(1) \}.$$

These are $$\Bbb Q$$-algebras (right?). I want to write these as quotients of polynomial rings. How do I go about finding the generators of these algebras, and then after having set up the homomorphism, determining its kernel (i.e. the ideal $$I$$ by which the polynomial ring $$R$$ is divided such that $$R/I\cong A$$ or $$R/I\cong B$$?) I heard of some procedure called the Gröbner basis.

It seems to me that this is explained nowhere. Could someone give a reference or explain to me how I would go about doing this?

• If you have a generating system $a_1,a_2,\dots$, then take $I:=\{f\in R[x_1,x_2,\dots] : f(a_1,a_2,\dots)=0\}$. Sep 7, 2020 at 11:15
• Of course, that is the definition of the kernel. Sep 7, 2020 at 11:19
• Well, yes. But looking from another perspective, $I$ collects all relations among the generators expressible in the language of rings that hold in the target ring. Sep 7, 2020 at 11:27

It's just hard in general. $$A$$ and $$B$$ are given as subalgebras satisfying certain conditions so you just have to figure out how to generate all elements satisfying those conditions, and that takes work.

Let's do $$B$$ first. Everything we're about to do works over an arbitrary field $$k$$. We know that $$f(0)$$ is the value of $$f(x) \bmod (x)$$ and similarly that $$f(1)$$ is the value of $$f(x) \bmod (x-1)$$. If we impose a condition that compares them then that condition only depends on the value of $$f(x) \bmod (x^2 - x)$$. We can write

$$f(x) = a + bx + (x^2 - x) g(x)$$

for some $$g$$, and then the condition we want is that $$f(0) = a = f(1) = a + b$$, hence that $$b = 0$$. So $$B$$ is the algebra of polynomials of the form $$a + (x^2 - x) g(x)$$. A set of generators is given by

$$y = x^2 - x, z = yx = x^3 - x^2$$

and we can see this as follows. We want to generate every polynomial of the form $$a + y g(x)$$. By taking linear combinations of $$\{ 1, y, z \}$$ we get all such polynomials where $$g$$ is linear. By adding $$y^2 = y(x^2 - x)$$ we get $$g$$ quadratic. By adding $$yz = y(x^3 - x^2)$$ we get $$g$$ cubic. And so forth.

As for relations, we have

$$z^2 = y^2 x^2 = y^2 (y + x) = y^3 + yz$$

and we can check that this generates the ideal of all relations by checking that $$B$$ is an integral extension of $$k[y]$$ by $$z$$ and then computing the minimal (monic, quadratic) polynomial of $$z$$ over $$k[y]$$, which is the above. Altogether we have that

$$\boxed{ B \cong k[y, z]/(z^2 - y^3 - yz) }.$$

A very similar argument shows that $$A$$ is the algebra of polynomials of the form $$a + x^2 g(x)$$ and then I invite you to check as an exercise very similar to the above that this algebra is generated by $$y = x^2, z = x^3$$ with relations generated by $$z^2 = y^3$$, so

$$\boxed{ A \cong k[y, z]/(z^2 - y^3) }.$$