# Intuition of quotient rings of polynomials

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?

• 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. Jun 26, 2023 at 16:52

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 $$V$$.
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: $$k[x,y]/I \simeq k[x,y]/J \times k[x,y]/K \simeq \Gamma(k) \times \Gamma(k)$$