Isomorphism $k[x,y]/(y-x^2)$ onto $k[x]$

Question

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

Answer 1

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 $\ker \varphi = (y - x^2)$.

Answer 2

Here's a detailed proof for the future readers.

Claim: If $K$ is a field, then $K[x,y]/(y-x^2) \cong K[x].$

Proof. Define $$\phi: K[x,y] \to K[x]$$ by $f(x,y) \mapsto f(x, x^2)$

It is easy to check that $\phi$ is a ring homomorphism. All that remains is to show that $\phi$ is surjective and that $ker \; \phi = (y-x^2)$.

As $K[x]$ is embedded in $K[x,y]$ as constant polynomials in $y$, we have that every $f(x,x^2)$ in $K[x]$ has a pre-image in $K[x,y]$. Thus, $\phi$ is surjective.

$$ker\; \phi = \{f(x,y) \in K[x,y]: f(x, x^2) = 0\}$$ Clearly, for any $g(x,y) \in K[x,y]$, we have $g(x,y) \cdot (y-x^2) \in ker \; \phi$.

Thus, $(y-x^2) \subseteq ker\; \phi$.

To show the other inclusion, let $p(x,y) \in ker\; \phi$. As $y-x^2$ is monic in $y$, we can divide $p$ by $y-x^2$. Thus, there exists $q(x,y)$ and $r(x,y)$ in $K[x,y]$ such that: $$p = (y-x^2)q(x,y)+r(x,y)$$ where degree of $r<1$ in $y$. This means that $r$ is a polynomial in $x$ only.

Now as $p \in ker\; \phi$, $\; p(x, x^2) = 0 \implies r(x) = 0$

Therefore, we have that $p = (y-x^2)q \implies p \in (y-x^2)$

Hence, $ker\; \phi \subseteq (y-x^2)$ and thus, $ker \; \phi = (y-x^2)$

$\therefore$ By the first isomorphism theorem, it follows that $K[x,y]/(y-x^2) \cong K[x]. \; \blacksquare$

Answer 3

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.