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This question was asked in my assignment on the course of algebraic geometry and I was not able make significant progress on it.

Question: Consider the algebraic subset given by $P_1= V(X^2 -Y)$. Show that $\mathbb{A}^1 \approx V(X^2-Y)$ and find 1 such isomorphism.

Here $\mathbb{A}^1$ means affine line.

Let $S$ be an arbitrary subset of $k[X_1,...,X_n]$. We set $V(S)= ${$x\in k^n | \forall P \in S, P(x)=0$}; ie the $x\in V(S)$ are the common zeroes of all the polynomials in $S$.

I am able to see what $V(S)$ will look like and the zero set is $V(S)= ${$(x,y)\in \mathbb{R}^2 | x^2-y=0$} but I am not able to understand how a function between the above set and $\mathbb{A}^1$ exists, much less how it will be isomorphism.

Can you please help me with proving this exercise?

BKFH
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  • The isomorphism is $(x,x^2) \in V(X^2-Y)\mapsto x \in k$, so the projection on the first variable. Draw it for $\mathbb{R}$, this will be very clear – julio_es_sui_glace Nov 25 '23 at 20:33
  • One way to do this (if you saw coordinate rings already) is to show there is an isomorphism of coordinate rings. This is just a repackaging of julio_es_sui_glace’s answer. – Shrugs Nov 25 '23 at 20:35
  • Many introductory algebraic geometry problems have already been asked and answered here on MSE. In the future, please take some time to look around for them - if you find something which answers your question, great! If not, it would be helpful to mention what you've looked at and why it's not enough in your question. This question, for instance, has been solved before in the linked duplicate. – KReiser Nov 26 '23 at 01:57
  • @KReiser Can you please give me some tips to search! – BKFH Nov 26 '23 at 15:54
  • Standard search engines can be helpful, if you put in the correct queries - a few key words from the problem and especially the book name and exercise number if your problem can be found in a textbook (sometimes your problem is in a textbook you're not aware of, which can be a fun surprise). There's also Approach0, a math-aware search engine - Meta announcement. The on-site search is iffy, but the related question suggestions are frequently better, so sometimes you can start typing up your problem and find something there. – KReiser Nov 28 '23 at 02:30

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One can represent the affine line as $k$ (they are isomorphic as varieties)., we just need to find an isomorphism (which is by definition polynomial) between $V(X^2-Y)= $ and $k$, (I don’t know how formal your proof needs to be and what your definitions are so you may need to adapt a bit)

Let us set $\phi:k \to V(X^2-Y)$, defined by $\phi(x) = (x,x^2)$. Of course it is polynomial and well defined. The injectivity is obvious, but for $(x,y) \in V(X^2-Y)$, we have $x^2 = y$, so $\phi(x) : (x,y)$, hence it is surjective.

Here you have your isomorphism!

julio_es_sui_glace
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  • so we don't have to prove this map k-algebra homomorphism or ring homomorphim? – BKFH Nov 25 '23 at 20:53
  • That is why I put the commentary in parenthesis, I don’t know how you defined things and how far you are, but you could try to work with quotient rings, it would be completely fine. – julio_es_sui_glace Nov 25 '23 at 20:58
  • I am reading from 1st chapter of Hartshorne's book. – BKFH Nov 25 '23 at 21:01
  • It has been some time since I've done algebraic geometry, so I won't go any more formal; you can try to adapt this reasoning to your level. Just write it down very slowly it should be ok – julio_es_sui_glace Nov 25 '23 at 21:36