4
$\begingroup$

Let $k$ be a field, $R=k[x_1,\ldots, x_n]$, $V$ an algebraic set and $I:=I(V)$ the ideal formed by the polynomials that vanish on all points of $V$. I am trying to get a firm intuition of what the quotient $R/I$ is.

I got told that this quotient is exactly the set of polynomial functions on $V$. To start with, I don't see why this is true. I mean, if I write down some examples I kind of see how this can be true but I don't have an intuition of why it happens; I hope someone can help me with this.

Now let's write down a couple of examples. For the first one, I'll choose $k=\mathbb{C}$ and $n=2$. I am therefore working in $\mathbb{A}_\mathbb{C}^2$. I will take $V= \{y=0\} \cong \mathbb{A}_{\mathbb{C}}^1$. The quotient $\mathbb{C}[x,y]/I\cong \mathbb{C}[x]$ and I can understand that the set $\mathbb{C}[x]$ is exactly the set of polynomial functions on $V=\mathbb{A}_{\mathbb{C}}^1$. However, if now I write $V'= \{y=1, y=-1\}$ this is like two instances of $\mathbb{A}_{\mathbb{C}}^1$. However, when taking the quotient $\mathbb{C}[x,y]/I(V')$ where $I(V')=(y^2-1)$, I get that this is again $\mathbb{C}[x]$, and I don't see that in this case this is the set of polynomial functions on two instances of the affine space. Where am I going wrong?

$\endgroup$
1
  • 1
    $\begingroup$ 1. Please put all math text inside math mode. To deal with special characters, one may escape them like so: $\{y=0\}$ produces $\{y=0\}$. 2. You have incorrectly computed the quotient: $\Bbb C[x,y]/(y^2-1)\cong \Bbb C[x] \times \Bbb C[x]$ by the Chinese remainder theorem. $\endgroup$
    – KReiser
    Jun 26, 2023 at 16:52

2 Answers 2

4
$\begingroup$

Let $f,g\in R$ and let $\overline{f}$ and $\overline{g}$ denote their images in the quotient. Then $\overline{f}=\overline{g}$ if and only if $f-g\in I$, in other words, if and only if $f-g$ vanishes on $V$. But this means $f$ and $g$ are equal as functions on $V$. The point is, you can have two different polynomials in $R$ give the same polynomial function on $V$, and you want to think of these as being equal, so that the quotient $R/I$ is exactly the polynomial functions on $V$.

$\endgroup$
3
$\begingroup$

I'd add to what @morrowmh said that to see it in a really algebraic way, you must use the quotient theorem on the ring morphism: $$\begin{array}{rcl} R & \longrightarrow & \Gamma(V)\\ f & \longmapsto & (x \in V \mapsto f(x)) \end{array}$$

Its kernel is exactly $I$, thus you get the iso $R/I \simeq \Gamma(V)$. You "kill" all polynomials that are the same on $V$

For your second example, you can easily see that $\Gamma(V') = \Gamma(k) \times \Gamma(k)$; just consider two copies of polynomial functions defined on the line. On the more algebraic part, you have $I = (y^2-1) = JK$ with $J = (y-1)$, $K=(y+1)$ since both of these ideal are coprime ($y+1 - (y-1) = 2$) by Chinese remainder the you get: $$k[x,y]/I \simeq k[x,y]/J \times k[x,y]/K \simeq \Gamma(k) \times \Gamma(k)$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .