Let the ring be $A=k[x^2,x^3]$ where $k$ is a field. Let $I$ be the ideal generated by $x^4$. I want to prove that there's only one ideal in $A/I$. For that I considered that in $k[x]$ the ideals that contain $x^4$ are exactly the ideal of lower power of $x$. So that leaves me the choices of $(x^3), (x^2), (x)$ but only the ideal of $(x^2)$ contains $x^4$ in A. So the only ideal in $A/I$ is exactly $(\bar{x^2})$.

Is this reasoning/result correct? Thanks!

  • $\begingroup$ What about the ideal $(x^3,x^4)$? $\endgroup$ – Mohan Jul 6 '18 at 17:39
  • $\begingroup$ @Arthur How so? There's something I'm not quite understanding. $\endgroup$ – Leo Jul 6 '18 at 17:42
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    $\begingroup$ Yes, but what have you tried? Any ideal of $A/I$ corresponds to an ideal of $A$ containing $I$. There are only finitely many of them and easy to write down with a little thought. $\endgroup$ – Mohan Jul 6 '18 at 18:46
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    $\begingroup$ @Mohan Finitely many? It seems like that depends on $k$. $\endgroup$ – rschwieb Jul 6 '18 at 18:54
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    $\begingroup$ @rschwieb No, I did not mean finitely many and was a mistake. $\endgroup$ – Mohan Jul 6 '18 at 21:50

I am doing more than just determining the ideals of $A/I$.

Note that $A\cong k[y,z]/\langle y^3-z^2\rangle=:B$ via the ring isomorphism $\varphi:B\to A$ sending $y\mapsto x^2$ and $z\mapsto x^3$. Under $\varphi$, the ideal $I$ of $A$ is associated to the ideal $\langle y^2,y^3-z^2\rangle/\langle y^3-z^2\rangle$ of $B$. That is, $$A/I \cong k[y,z]/\langle y^2,y^3-z^2\rangle=:C\,,$$ which is a $4$-dimensional vector space over $k$ with basis $\{1,\bar{y},\bar{z},\bar{y}\bar{z}\}$, where $\bar{y}$ and $\bar{z}$ are the images of $y$ and $z$, respectively, under the canonical projection $k[y,z]\to k[y,z]/\langle y^2,y^3-z^2\rangle$.

Observe that $\bar{y}^2=0$ and $\bar{z}^2=0$ in $C$. Hence, $C$ is in fact isomorphic as a ring to $$k[u,v]/\langle u^2,v^2\rangle\cong k[u]/\langle u^2\rangle \otimes k[v]/\langle v^2\rangle\,,$$ where $\otimes$ is the tensor product of algebras over $k$. Note that $C$ has many ideals: the zero ideal $0$, the whole ring $C$ itself, the primitive ideals $\langle \bar{y}+\alpha\bar{z}+\beta \bar{y}\bar{z}\rangle$ with $\alpha,\beta \in k$, the primitive ideals $\langle \bar{z}+\gamma \bar{y}\bar{z}\rangle$ with $\gamma\in k$, the primitive ideal $\langle \bar{y}\bar{z}\rangle$, and the unique maximal ideal $\langle \bar{y},\bar{z}\rangle$. If we translate this list back to ideals of $A/I$, then all the ideals of $A/I$ are:

  • trivial ideals: $0$ and $A/I$;
  • primitive ideals: $\langle \overline{x^2}+\alpha\overline{x^3}+\beta\overline{x^5}\rangle$, $\langle \overline{x^3}+\gamma\overline{x^5}\rangle$, and $\langle \overline{x^5}\rangle$, where $\alpha,\beta,\gamma \in k$;
  • maximal ideal: $\langle \overline{x^2},\overline{x^3}\rangle$.

Here, $\overline{x^2}$, $\overline{x^3}$, and $\overline{x^5}$ are the images of $x^2$, $x^3$, and $x^5$ under the canonical projection $A\to A/I$. You can say that $A/I$ is a local ring.

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    $\begingroup$ Maybe it's useful to notice that $\langle y^2,y^3-z^2\rangle=\langle y^2,z^2\rangle$. This way there is no need to introduce two more variables $u,v$. $\endgroup$ – user26857 Jul 6 '18 at 22:16

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