# Difference between ideal-theoretic and set-theoretic definitions of varieties

While studying some algebraic geometry and more specifically secant varieties of Segre embeddings, I encountered a problem. I understand that every variety has a defining ideal such that the variety is the zero set of the ideal. However, I think I may not understand the notion of defining a variety set-theoretically.

As an example, consider the $x$-axis in $\mathbb{R}^2$. I understand it is the zero-set of the polynomial $y=0$, and that it's defining ideal is $\langle y \rangle$.

$\textbf{Question 1: }$ Am I correct in saying that the $x$-axis is generated set-theoretically by $\{y\}$ and ideal-theoretically by $\langle y \rangle$?

If this is the case, the following confuses me. I have seen papers (for example https://arxiv.org/abs/1104.1776) claiming that they defined a certain variety (in this case Sec$_4(\mathbb{P}^3\times \mathbb{P}^3\times \mathbb{P}^3$)) set-theoretically.

$\textbf{Question 2: }$ In this case, what stops me from defining the defining ideal of the variety as the ideal generated by the set of polynomials used for the set-theoretic definition?

I feel this should be possible, but then I would not understand why the above paper only works on a set-theoretic level and why in fact there is a difference.

I would greatly appreciate any help!

Schemes are locally ringed spaces, which means that in order to specify a scheme, one needs to provide a topological space and structure sheaf. Saying that one has specified a subscheme set-theoretically means that one has given the topological space, but not the structure sheaf. There are many possible choices for the structure sheaf that one could make - for your example of the $x$-axis inside $\Bbb A^2$, it may be set-theoretically specified as $V(y^n)$ for any positive integer $n$, and it's clear that each of these structures are pair-wise non-isomorphic.
• Thank you for your comment! I do not know anything about schemes so I will look into them. Am I correct in interpretating your first answer as saying that there is no natural way to obtain the ideal $\langle y \rangle$ starting from the set $\{ y^4\}$ for example. In my case this was the case because I 'naively' started with exactly $\{ y \}$. – MisterT Nov 11 '17 at 20:46
• What do you mean when you're talking about $\{y^4\}$? – KReiser Nov 11 '17 at 20:52
• I mean that one can define the $x$-axis set theoretically as the zero set by the polynomial $y^4$. While asking the question I assumed there was a simple way to obtain the defining ideal of the $x$-axis from this, by just taking $y^4$ as a generator of the ideal. However, this is clearly not the case – MisterT Nov 11 '17 at 20:54
• There's a simple way to specify the defining ideal like that, and the most usual way is as $V(y^4)$ (or $V(I)$ for some specified ideal $I$). I was just confused about your notation. – KReiser Nov 11 '17 at 21:21