This is related to an exercise in Hartshorne's book which people have previously asked about, but I feel that this particular point hasn't been answered in a very simple way.

Let $k$ be an algebraically closed field. Let $k[x]$ be the ring of polynomials of one variable over $k$ and let $k[x,y]$ be the ring of polynomials of two variables over $k$. How can we see the isomorphism $$ k[x] \simeq k[x,y]/\langle y-x^2 \rangle, $$ where $\langle y-x^2 \rangle$ is the ideal in $k[x,y]$ generated by $y-x^2$?

  • $\begingroup$ are you asking for a explicit construction of such isomorphism or just how can you intuitively notice that these two rings are isomorphic? $\endgroup$ – JayTuma Aug 11 '18 at 15:23
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    $\begingroup$ No need to assume that $k$ is algebraically closed. $\endgroup$ – lhf Aug 11 '18 at 15:24
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    $\begingroup$ How about this answer, which seems to be very direct? $\endgroup$ – user99914 Aug 11 '18 at 15:25


$$\phi: k[x,y]\to k[x],\;x\to x,\;\;y\to x^2,\text{ and expand accordingly}$$

or if you prefer: $\phi f(x,y):=f(x,x^2)$. Now check stuff.

  • $\begingroup$ It's easy to see that $\phi$ is indeed a homomorphism. Any tips on how to see it's an isomorphism too? $\endgroup$ – Matias Heikkilä Aug 11 '18 at 15:30
  • $\begingroup$ @Pedro ¿Ahorrando signos, Pedro? La verdad no le veo mucho caso. Gracias, igual. $\endgroup$ – DonAntonio Aug 11 '18 at 15:31
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    $\begingroup$ @MatiasHeikkilä Oh, it is not an isomorphism, of course. What is its kernel? Well, now apply the first isomorphism theorem (for rings).\ $\endgroup$ – DonAntonio Aug 11 '18 at 15:32
  • $\begingroup$ ah, silly me. Thank you! $\endgroup$ – Matias Heikkilä Aug 11 '18 at 15:35
  • $\begingroup$ For the future reader: Details on the kernel of $\phi$ can be found here. $\endgroup$ – Matias Heikkilä Aug 11 '18 at 15:58

Hint: Let $D$ be a domain and $a \in D$. Then $D[y] \to D$ given by $f(y) \mapsto f(a)$ is a surjective ring homomorphism with kernel $\langle y-a \rangle$ because $f(y)=(y-a)q(y)+f(a)$. Apply this to $D=k[x]$.


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