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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?

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    $\begingroup$ If you have a generating system $a_1,a_2,\dots$, then take $I:=\{f\in R[x_1,x_2,\dots] : f(a_1,a_2,\dots)=0\}$. $\endgroup$
    – Berci
    Sep 7, 2020 at 11:15
  • $\begingroup$ Of course, that is the definition of the kernel. $\endgroup$ Sep 7, 2020 at 11:19
  • $\begingroup$ 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. $\endgroup$
    – Berci
    Sep 7, 2020 at 11:27

1 Answer 1

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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

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

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