Coordinate ring of an affine algebraic set - k[A^n]/I(V)

[Math Amateur]

I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

I need someone to help me to fully understand the reasoning/analysis behind the statements in the Example on Page 661 of D&F regarding the coordinate ring of an affine algebraic set $$V \subseteq \mathbb{A}^n$$ ... ...

On page 661 (in Section 15.1) of D&F we find the following text and example (I am specifically focused on the Example):
https://www.physicsforums.com/attachments/4752
In the above text, in the Example, we find the following:

" ... ... In the quotient ring $$\mathbb{R} [V]$$ we have $$\overline{x} \overline{y} = 1$$ so $$\mathbb{R} [V] \cong \mathbb{R} [ x, 1/x ]$$. ... ... "

Could someone please explain rigorously and formally and, preferably in simple steps, exactly how/why $$\mathbb{R} [V] \cong \mathbb{R} [ x, 1/x ]$$ ... ... ?

Further, in the above text, D&F state " ... $$\overline{x} \overline{y} = 1$$ ... " ... ... BUT ... ... shouldn't this read " ... $$\overline{x} \overline{y} = \overline{1}$$ ... "

Hope someone can help with the above issues ...

Peter

 

[mathbalarka]

I'll answer the question first :

$\Bbb R[x, y]/(xy - 1)$ is the same thing as $\Bbb R[x, 1/x]$. There is nothing much to prove. Note that $\Bbb R[x, 1/x]$ is the ring of formal polynomials of $x$ and $1/x$. That is, it's NOT a polynomial ring. This is a very important note to bear in mind. A polynomial ring over $\Bbb R$ is of the form $\Bbb R[x_1, x_2, \cdots, x_n]$ where $x_i$ are $R$-algebraically independent transcendentals. $\Bbb R[x, 1/x]$ is what you get by treating $x$ and $1/x$ as symbols and building polynomials with them.

That said, the isomorphism is $f : \Bbb R[x, y]/(xy - 1) \to \Bbb R[X, 1/X]$ given by $1 \pmod{(xy-1)} \mapsto 1$, $x \pmod{(xy-1)}\mapsto X$ and $y \pmod{(xy-1)} \mapsto 1/X$ and the rest by extending to all of the ring. This is a homomorphism, as $f(xy \pmod{(xy-1)}) = f(1 \pmod{(xy-1)}) = 1$ and $f(x \pmod{(xy-1)})f(y \pmod{(xy-1)}) = X \cdot 1/X = 1$. You can prove this is an isomorphism by noting that this is surjective and has zero kernel. I'll leave that to you to verify.

---

Since you are reading about the coordinate ring, I'll give a few high-brow comments :

If $V \subset \Bbb A^n$ is an affine variety, $\mathcal{I} = \mathcal{I}(V)$ the ideal corresponding to $V$ (ideal of _all_ the functions vanishing identically on $V$). Then coordinate ring $k[V] \cong k[x_1, \cdots, x_n]/\mathcal{I}$ is actually the ring of functions on $V$.

That is, consider an element $f(x_1, \cdots, x_n) \pmod{\mathcal{I}} \in k[V]$. If you evaluate this thing on some point $(a_1, \cdots, a_n) \in V$ on the variety $V$, then you get back the value $f(a_1, \cdots, a_n)$ (regardless of the choice of the thing that's dangling inside the "mod" sign. This is because any polynomial in $\mathcal{I}$ evaluates to $0$ on $V$). Thus, an element of $k[V]$, evaluated at points of $V$, gives rise to a function $V \to k$. The opposite direction is also true.

Thus, we conclude that $k[V]$ is actually ring of functions $V \to k$ on $V$ (multiplication and addition of two functions $f, g : V \to k$ is defined by multiplication and addition of the images in $k$).

This is an immensely important notion in algebraic geometry. When you learn about Zariski topology next, you'll see that you can localize the ring $k[V]$ at the open sets of your affine variety to form a sheaf on the variety. Zariski topology is completely useless as a topology, but it's great when you want to do sheaf theory on varieties. This is a notion motivated from complex analysis, where you study sheaf of holomorphic functions on complex manifolds. That gives a lot of information about line bundles over your manifold, say.

To give a taste of how powerful this sheaf theory is, let $V$ be an affine variety and $k[V]$ be a coordinate ring. It's known that points in $V$ are in bijective correspondence with maximal ideals of $k[V]$ (this is a weak version of Nullstellensatz, as you'll learn later). In fact, you can give $V$ and the set of maximal ideals of $k[V]$ both a topology (the Zariski topology, of course), in which case the bijection correspondence will become a homeomorphism between $V$ and $\text{maxSpec} \, k[V]$ (the collection of maximal ideals). Hence, you can recover the whole variety just from the maximal ideals of the coordinate ring.

 

[Math Amateur]

I'll answer the question first :

$\Bbb R[x, y]/(xy - 1)$ is the same thing as $\Bbb R[x, 1/x]$. There is nothing much to prove. Note that $\Bbb R[x, 1/x]$ is the ring of formal polynomials of $x$ and $1/x$. That is, it's NOT a polynomial ring. This is a very important note to bear in mind. A polynomial ring over $\Bbb R$ is of the form $\Bbb R[x_1, x_2, \cdots, x_n]$ where $x_i$ are $R$-algebraically independent transcendentals. $\Bbb R[x, 1/x]$ is what you get by treating $x$ and $1/x$ as symbols and building polynomials with them.

That said, the isomorphism is $f : \Bbb R[x, y]/(xy - 1) \to \Bbb R[X, 1/X]$ given by $1 \pmod{(xy-1)} \mapsto 1$, $x \pmod{(xy-1)}\mapsto X$ and $y \pmod{(xy-1)} \mapsto 1/X$ and the rest by extending to all of the ring. This is a homomorphism, as $f(xy \pmod{(xy-1)}) = f(1 \pmod{(xy-1)}) = 1$ and $f(x \pmod{(xy-1)})f(y \pmod{(xy-1)}) = X \cdot 1/X = 1$. You can prove this is an isomorphism by noting that this is surjective and has zero kernel. I'll leave that to you to verify.

---

Since you are reading about the coordinate ring, I'll give a few high-brow comments :

If $V \subset \Bbb A^n$ is an affine variety, $\mathcal{I} = \mathcal{I}(V)$ the ideal corresponding to $V$ (ideal of _all_ the functions vanishing identically on $V$). Then coordinate ring $k[V] \cong k[x_1, \cdots, x_n]/\mathcal{I}$ is actually the ring of functions on $V$.

That is, consider an element $f(x_1, \cdots, x_n) \pmod{\mathcal{I}} \in k[V]$. If you evaluate this thing on some point $(a_1, \cdots, a_n) \in V$ on the variety $V$, then you get back the value $f(a_1, \cdots, a_n)$ (regardless of the choice of the thing that's dangling inside the "mod" sign. This is because any polynomial in $\mathcal{I}$ evaluates to $0$ on $V$). Thus, an element of $k[V]$, evaluated at points of $V$, gives rise to a function $V \to k$. The opposite direction is also true.

Thus, we conclude that $k[V]$ is actually ring of functions $V \to k$ on $V$ (multiplication and addition of two functions $f, g : V \to k$ is defined by multiplication and addition of the images in $k$).

This is an immensely important notion in algebraic geometry. When you learn about Zariski topology next, you'll see that you can localize the ring $k[V]$ at the open sets of your affine variety to form a sheaf on the variety. Zariski topology is completely useless as a topology, but it's great when you want to do sheaf theory on varieties. This is a notion motivated from complex analysis, where you study sheaf of holomorphic functions on complex manifolds. That gives a lot of information about line bundles over your manifold, say.

To give a taste of how powerful this sheaf theory is, let $V$ be an affine variety and $k[V]$ be a coordinate ring. It's known that points in $V$ are in bijective correspondence with maximal ideals of $k[V]$ (this is a weak version of Nullstellensatz, as you'll learn later). In fact, you can give $V$ and the set of maximal ideals of $k[V]$ both a topology (the Zariski topology, of course), in which case the bijection correspondence will become a homeomorphism between $V$ and $\text{maxSpec} \, k[V]$ (the collection of maximal ideals). Hence, you can recover the whole variety just from the maximal ideals of the coordinate ring.

Hi Mathbalarka

Thanks for an informative and interesting post ... you seem to have a truly impressive knowledge of algebraic geometry ... hope I can follow you ...

I am still thinking and reflecting over what you have said ...

Peter

 

[mathbalarka]

you seem to have a truly impressive knowledge of algebraic geometry

Nah, I have just recently finished studying the first few chapters of Atiyah-Macdonald. Really don't know any algebraic geometry, but I'd have to learn it soon.