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Let $D$ be a UFD over a field $k$ of characteristic zero. Assume that $w$ is algebraic over $D$. Denote $R=D[w]$.

Observe that $R$ is not necessarily a UFD.

Can one find an example in which $R$ has no prime elements at all?

My example: $D=k[x^2]$, $w=x^3$ (its minimal polynomial over $k[x^2]$ is of degree $2$. Notice that $x^3$ is integral over $D$), $R=k[x^2][x^3]$; is it true that $k[x^2,x^3]$ has no prime elements? Of course, as a Noetherian ring (or more generally, as a ring which satisfies ACCP), $R$ has irreducible elements, for example, $x^2$ and $x^3$, but these are not prime elements since $(x^2)(x^2)(x^2)=(x^3)(x^3)$.

See also this question.

Remark: If $w$ is transcendental over $D$, then $R=D[w]$ is a UFD, as a polynomial ring (in one variable) over a UFD.

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  • $\begingroup$ "is it true that $k[x^2,x^3]$ has no prime elements?" Yes, if $k$ is algebraically closed. However, $x^2+1$ is prime in $\mathbb R[x^2,x^3]$. $\endgroup$
    – user26857
    May 12, 2017 at 6:57
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    $\begingroup$ @user26857, thanks for your comment! I see why $x^2+1$ is irreducible in $\mathbb{R}[x^2,x^3]$. As for primality: $x^2+1$ is prime if and only if $(x^2+1)$ is a prime ideal if and only if $A:=\mathbb{R}[x^2,x^3]/(x^2+1)$ is an integral domain. If I am not wrong $A$ is isomorphic to $\mathbb{R}[x]$ which is, of course, an integral domain. $\endgroup$
    – user237522
    May 14, 2017 at 21:53
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    $\begingroup$ Unfortunately you are wrong: $A\simeq\mathbb R[x]/(x^2+1)$. This is the point of this example: if change $\mathbb R$ by $\mathbb C$ then the primality is gone. $\endgroup$
    – user26857
    May 15, 2017 at 6:39
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    $\begingroup$ Thanks for your correction that $A \simeq \mathbb{C}[x]$. $\endgroup$
    – user237522
    May 15, 2017 at 12:38
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    $\begingroup$ Do you mean $A\simeq\mathbb C$? $\endgroup$
    – user26857
    May 15, 2017 at 12:40

1 Answer 1

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A prime element is nothing else but a non-zero principal prime ideal.

$\mathbb C[x^2,x^3] \cong \mathbb C[u,v]/(u^2-v^3)$ is one-dimensional and the non-zero prime ideals are given by $(u,v)$ and $(u-U,v-V)$ where $U,V \in k \setminus \{0\}$ with $U^2=V^3$.

$(u,v)$ is not principal, because the curve is not regular at that point. All other points are regular, i.e. the corresponding prime ideals are locally principal. We still have to check whether they are principal.

Of course by rescaling, we find an automorphism which lets us assume $U=V=1$, i.e. we only have to show that $(u-1,v-1) \subset \mathbb C[u,v]/(u^2-v^3)$ is not principal. In other words we have to show that $(x^2-1,x^3-1) \subset \mathbb C[x^2,x^3]$ is not principal. This is easy because both generators are irreducible by the virtue of the absence of linear polynomials. Thus the only principal ideal, that contains $x^2-1$ and $x^3-1$ is the ideal $(1)$, but we already know that $(x^2-1,x^3-1) \neq (1)$, because it is a prime ideal.

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  • $\begingroup$ Nice answer, thanks! Please, do you know what happens in case $D=k[x,y]$? (or $D=\mathbb{C}[x,y]$, if it helps that the base field is algebraically closed). It probably depends on the specific algebraic element $w$ (more accurately, it probably depends on its minimal polynomial). $\endgroup$
    – user237522
    May 11, 2017 at 13:49
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    $\begingroup$ Just a note:I only chose the field to be algebraically closed to ensure that rescaling is always possible, i.e. it was just to make the argument more smooth. The counter example should work over any field. For $D=k[x,y]$ on has to make sure that $y$ is algebraic over $k[x,w]$ and $x$ is algebraic over $k[y,w]$, because otherwise $k[x,y,w]$ is a polynomial ring over a domain, i.e. the indeterminate is a prime element. Maybe $k[x^2,y^2,xy]$ (or $k[x,y,\sqrt{xy}]$ of you like it that way) works as an example here. $\endgroup$
    – MooS
    May 11, 2017 at 13:52
  • $\begingroup$ Thanks for the clarification about the base field! Concerning $k[x,y]$: Please do you think that there exists a 'general' answer in terms of the coefficients of $f$, the minimal polynomial of $w$ over $D$? In your example, $f(t)=t^2-xy$ is the minimal polynomial of $w=\sqrt{xy}$. Perhaps if $f(0)$ is irreducible in $k[x,y]$ cannot serve as an example. $\endgroup$
    – user237522
    May 11, 2017 at 14:06
  • $\begingroup$ I cannot imagine that (2) is anyhow relevant, because the number of coefficients can be very large, but the length of a regular sequence is pretty bounded. isn't it? Maybe (1) plays a role though. But you should examine these questions for $D=k[x]$ first, shouldn't you? Even in this case, we have only given one example so far, we have not obtained any general results on the existence of prime elements in $k[x,w]$ where $w$ is algebraic over $k[x]$. Anyway this is indeed worth another question. $\endgroup$
    – MooS
    May 11, 2017 at 14:11
  • $\begingroup$ Thanks for your comment! (I deleted: "Perhaps an answer depends on: (1) $w$ is integral or not. (2) the coefficients of $f$ forms a regular sequence"). $\endgroup$
    – user237522
    May 11, 2017 at 14:15

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