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In the instance of a line $L$ intersecting with a parabola I need to find out if the intersection is a dense subset. Here's my work / question so far:

For another point $P$ on the parabola not in the intersection of $L$, is the entire complement of another variety which contains that point counted as a neighborhood or just the connected part around the point? That is whenever I look at a circle $C$ centered at $P$ which doesn't have the intersection of the line within the radius then is the complementary set of $C$ which has two parts- one which has the point in it and the other with the line intersecting the parabola counted as a neighborhood around $P$ which contains the intersection of the line? The text 'An Invitation to Algebraic Geometry' does say that the open sets are very large so can't that just mean that the set is dense along with that the two parts in and outside of the circular variety are connected, because they can not really be complementary sets of any polynomials!

  • Can we clarify the question? You are with an arbitrary line $L$ and a parabola $X$ - defined over an (algebraically closed) field $k$ and embedded in $\mathbb{A}_k^2$? You are then asking when $X \cap L$ is dense in $X$, (or in $L$?, or in $\mathbb{A}_k^2$?) – Mummy the turkey Jun 28 '20 at 00:33
  • I am only asking if it's dense in X. –  Jun 28 '20 at 00:40
  • The intersection is a closed proper subset of the parabola, so it can't be dense. But I can't really make sense of what you wrote to see where you're confused. –  Jun 28 '20 at 00:40

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From your question you seem to be slightly confused about what the Zariski topology looks like. We do not want to look at "open balls of radius $r$" like we would in the usual euclidean topology, instead we want to be looking at zero sets of polynomials.

Recall that a set $Y \subset X$ is dense if and only if $\bar{Y} = X$. Now, the line $L$ and the parabola $X$ are both closed subsets of $\mathbb{A}^2$ since they are the vanishing sets of some polynomials $y - mx - c$ and $y - x^2 - ax - b$ respectively. Thus the intersection $L \cap X$ is a proper closed subset of $X$ hence not dense.

Mummy the turkey
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  • Would the intersection be dense in the relative topology on the parabola ? –  Jun 28 '20 at 00:44
  • @dabofskateboarding no, the set $L$ is closed in $\mathbb{A}^2$ so $L \cap X$ is a closed subset of $X$ in the subspace topology (which is the Zariski topology in any case). – Mummy the turkey Jun 28 '20 at 00:47
  • @dabofskateboarding Let $X$ be a variety. If a point $Q$ is in the Zariski closure of a subset $Y \subset X$ then any $f \in k[X]$ vanishing on $Y$ must also vanish at $Q$ – Mummy the turkey Jun 28 '20 at 01:25