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Let $Y=V(x^2-y)$ be the affine variety $y=x^2$. We know that this is an irreducible variety, so $I(Y)$ is a prime ideal of $K[x,y]$ with affine coordinate ring $K[x,y]/I(Y)$, where since this is a radical ideal, Hilbert's nullstellensatz tells us that $I(Y)=(x^2-y)$, so we really have: $$A(Y)=K[x,y]/(x^2-y).$$

I just want to deduce that this is a polynomial ring in one variable. I could explicitly use the first isomorphism theorem to see $K[x,y]/(x^2-y)=K[x]$, but I should be able to do this with less work?

Actually I guess now that I think about it, my stuff (prior to edit) drew out what I really want to ask:

If $Y$ is an irreducible variety, then I know that the dimension of $Y$ is the same as the dimension of $A(Y)=K[X_1,\cdots, X_n]/I(Y)$, so I know in my case that $A(Y)$ is $1$-dimensional, and since $I(Y)$ is prime, I know that $A(Y)$ is an integral domain.

Can I immediately deduce that $A(Y)$ is a one variable polynomial ring over $K$?

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    Are you saying $Y=V(f)$ where $f\in k[x,y]$ is an irreducible polynomial has $A(Y)$ a polynomial ring? – 2017-02-06
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    @Mohan That's true isn't it? And in particular it is an integral domain here, since $I(Y)$ is prime? – 2017-02-06
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    It sounds like your question basically boils down to: if $A$ is an integral domain of transcendence degree $1$ over $K$, is $A$ isomorphic to $K[X]$? Or am I missing something? – 2017-02-06
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    @AlexMathers That sounds like the correct interpretation, and sounds true to me at this point. We send the single transcendental element to $X$, which is transcendental over $K$? (assuming they are over the same field, which here they are) – 2017-02-06
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    Take $f=xy-1$. Can you see why $A(Y)$ is not a polynomial ring in one variable? – 2017-02-06
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    @Mohan Since this is $K[x,x^{-1}]$? I am confused now. It seems that the polynomial is irreducible and prime, so this does give an irreducible affine variety. Which gives us something that looks 'disconnected' (but isn't mathematically, since in the Zariski, the two branches are not by themselves closed). We obtain a ring of fractions like $\langle x\rangle^{-1}K[X]$? (Localization at the ideal generated by $x$) – 2017-02-06
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    (Can't edit comment, but I shouldn't have used 'Ring of fractions' there, ignore those words) – 2017-02-06
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    I guess I cannot see *what* it is that makes it not a polynomial ring in one variable), although I can show it in this case. – 2017-02-06
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    $K[x,y]/f$ for $f$ irreducible is rarely isomorphic to $K[x]$. In fact for $K=\mathbb{C}$ (even more general), a theorem of Abhyankar-Moh says, if it is isomorphic, then after an automorphism of $K[x,y]$, $f=x$. Most of the time it is not. – 2017-02-08

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This answer from the book Elementray algebraic geometry, Klaus Hulek p.39 $Y=\{(x,y)\in \mathbb A^n_K;y-x^2=0 \}$, then $K[Y]=K[x,y]/(y-x^2)\cong K[x,x^2]=K[x] \cong K[\mathbb A^1_K] $

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Affine curves don't have to lines, or even subsets of quotients of lines.

For example, consider $Y$ to be an elliptic curve (minus a point at infinity).

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    I feel like you are hinting that it is a polynomial ring in one variable only in the case that it appears as a line. But in my case it is not a line, and still has one variable coordinate ring? I can't see where I wrote anything about lines – 2017-02-06