Prime ideals of $k[t^2,t^3]$

Question

I am trying to find a way to describe all prime ideals of $k[t^2,t^3]$, however I don't know how to even get started to find them (here $k$ is a field)

Are there any easy tricks to find/characterize them?

Answer 1

If you consider $R=k[x]$ and $$k[t^2,t^3]\cong R[y]/\langle y^2-x^3\rangle$$ then you are looking for prime ideals of $R[y]$ that include $y^2-x^3.$

Since $y^2-x^3$ is monic, we can find our ideal, $I,$ by knowing for which $p(x),q(x)\in R$ we have $p(x)+q(x)y\in I.$

But if $p(x)+q(x)y\in I,$ then $$p^2-x^3q^2=(p+qy)(p-qy)+q^2(y^2-x^3)\in I.$$

Also, if $q(x)\neq 0,$ then $p^2-x^3q^2\neq 0,$ since it is the difference between a polynomial of odd and even degree.

So if there are any elements of $I$ which are not multiples of $y^2-x^3,$ we must have $I\cap R\neq\{0\}.$ So $I\cap R=\langle f(x)\rangle$ for some irreducible polynomial $f.$

So $$\langle y^2-x^3,f(x)\rangle\subseteq I.$$

If there are any other elements of $I,$ there must be $p,q\in k[x]$ with $q\neq 0$ and $\deg p,q<\deg f$ and $p+qy\in I.$ But then $f\mid p^2-x^3q^2.$

Solving $fu+qv=1$ in $k[x],$ you get $$f(fu^2+2quv)+q^2v^2=1$$ and $$f\mid p^2v^2-x^3$$

This means that $y^2-p^2v^2=(y-pv)(y+pv)\in I,$ so one of $y\pm pv\in I.$ This would mean $I=\langle y+r,f(x)\rangle$ where $r\in R$ has $f(x)\mid r^2-x^3.$

On the other hand, $K=R/\langle f\rangle$ is a field, and $\langle x^3\rangle\in K$ is a square if and only if $\langle x\rangle\in K$ is a square.

So the non-zero prime ideals of $k[t^2,t^3]$ are determined by an irreducible $f\in k[x].$

1. For each $r\in k[x]$ of degree $<\deg f$ such that $f(x)\mid r^2-x,$ then $\langle f(t^2),t^3-r^3(t^2)\rangle$ is a prime ideal. This is a case where $f(t^2)$ is not irreducible in $k[t].$ $$f(t^2)\mid r^2(t^2)-t^2=(r(t^2)-t)(r(t^2)+t).$$
2. If there is no such $r,$ then $\langle f(t^2)\rangle$ is a prime ideal (see Lemma below) in $k[t^2,t^3].$

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In the first case, the ideal corresponds to the ideal $\langle \gcd(f(t^2),r(t^2)-t)\rangle$ in $k[t].$ This is a prime ideal because $f$ irreducible means $f(t^2)$ can have at most two irreducible factors.

In the second case, the ideal corresponds to $\langle f(t^2)\rangle$ in $k[t].$

So all prime ideals in $k[t^2,t^3]$ are intersections with prime ideals in $k[t].$

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Lemma: If $f(x)\in k[x]$ is irreducible and $f(x^2)$ is not, then there is an $r(x)\in k[x]$ with $f(x)\mid r^2(x)-x.$

Proof: $f(x^2)$ not being irreducible means there must be an irreducible factor $g$ with $\deg g\leq deg f.$

If $\alpha$ is a root of $g(x)$ then $\alpha^2$ is a root of $f(x).$ We clearly have $k[\alpha^2]\subseteq k[\alpha].$ But $$[k[\alpha]:k]=\deg g\leq\deg f= [k[\alpha^2]:k].$$ So the fields are identical, so $\alpha\in k[\alpha^2],$ and so $\alpha=r(\alpha^2)$ for some polynomial $r(x).$

So: $r^2(\alpha^2)=\alpha^2.$ Since $f$ is the minimal polynomial for $\alpha^2,$ then $f(x)\mid r^2(x)-x.$

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We can always find $r$ with $\deg r<\deg f.$ If the characteristic of $k$ is $2,$ there is only one such $r,$ otherwise there are $2.$

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Example: $k=\mathbb Q$ and $f(x)=1+x+x^2,$ then $r(x)=\pm(1+x),$ and we get two prime ideals:

$$ \begin{align} I_{\pm}&= \langle 1+t^2+t^4,t^3\pm(1+t^2)^3\rangle\\&= \langle 1+t^2+t^4,t^3\mp1\rangle \end{align} $$

These correspond to the ideals $\langle 1\pm t+t^2\rangle$ in $\mathbb Q[t].$

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Possibly easier approach, but I haven’t finished it

It might be easier to consider $I\subseteq k[t^2,t^3],$ and then realize that $I_2=I\cap k[t^2]$ is prime in $k[t^2]$ and $I_3=I\cap k[t^3]$ is prime in $k[t^3].$ Then there are irreducible (or zero) $f,g\in k[x]$ such that:

$$I_2+I_3\subseteq \langle f(t^2),g(t^3)\rangle\subseteq I.$$

Since $g(t^3)g(-t^3)\in I\cap k[t^2],$ and thus we require $f(t^2)\mid g(t^3) g(-t^3).$

From this, perhaps we can show this ideal is all of $I?$