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Perhaps this is very easy but I would like to get a proof of isomorphism $k[x,y]/(y-x^2) \cong k[x]$.

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    By the first isomorphism theorem, what you need is a homomorphism from $k[x,y]$ onto $k[x]$ with kernel $(y-x^2)$. Can you find such a thing?2012-03-29
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    Yes I tried, but there is an informal proof. I need a formal one though..2012-03-29
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    What? There is nothing informal about the approach I outlined.2012-03-29
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    I'm pretty sure that this has come up before. I'll look around. In any event, you seem to be able to find a homomorphism $k[x, y]/(y - x^2) \to k[x]$. You could use Euclidean division to check that the kernel is what you think it is. An easier method is to define an inverse. Where should $f(x) \in k[x]$ be sent?2012-03-29
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    @ChrisEagle Is $p(x,y) \mapsto p(x,x^2)$ the homomorphism we're looking for?2012-03-29
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    @MattN. You beat me by few seconds :) $(y-x^2) \mapsto (x^2 - x^2) = 0$2012-03-29
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    @J.D. : ) (some more characters)2012-03-29
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    I think we need to know what is the inverse now...2012-03-29
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    @AliBleybel Well, you have an inclusion $k[x] \to k[x, y]$. You could compose that with the quotient homomorphism and see what happens.2012-03-29

2 Answers 2

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Consider the ring of polynomials in two variables over a field $k$ which we call $k[x,y]$. Then what you can do now is you can consider $k[x,y]$ as $\big(k[x]\big)[y]$; you just simply collect powers of $y$ in a polynomial in $k[x,y]$.

Now you have the evaluation map $\varphi$ that sends $y$ to $x^2$ that is constant on the coefficients (in this case the coefficients are in $k[x]$) , so that $\varphi$ extends to a ring homomorphism from $k[x,y]$ to $k[x,x^2]$. Furthermore, we see that $k[x]$ is not different from $k[x,x^2]$ because $k[x]$ and $k[x,x^2]$ are respectively by definition the smallest rings containing $k$ and the set $\{x\}$, and $k$ and the set $\{x,x^2\}$. Therefore since you know now

$$\varphi: k[x,y] \longrightarrow\!\!\!\!\!\!\!\!\to k[x]$$

the first isomorphism tells you that $k[x,y]/\ker \varphi \cong k[x]$, where $\ker \varphi = (y - x^2).$

Exercise: Prove that $\ker \varphi = (y - x^2)$.

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Can't this also be seen geometrically? This may or may not be of interest to the OP, but I think it's worth posting. The only downside is that I think I need $k$ algebraically closed and of characteristic $0$.

Recall that coordinate rings of affine varieties are isomorphic iff the varieties they cut out are. Next, notice that $k[x,y]/ \langle y-x^2 \rangle$ is the coordinate ring of the usual parabola centered at the origin in $k^2$ and $k[x] \simeq k[x]/\langle 0 \rangle \simeq k[x,y]/\langle y \rangle$ is the coordinate ring of a copy of $k$. We can view this copy of $k$ as embedded into the plane, in which case the varieties are visually isomorphic, but the correspondence is given explicitly by $x \mapsto (x,x^2)$ or equivalently $(x,y^2) \mapsto x$, as others have noticed in different language.