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Let us take the ideal $I:= \langle xy-z,x^5-z^3 \rangle $ of the ring $\Bbb{C}[x,y,z].$ We want to find if this ideal is prime.

My thoughts: We define $f:=xy-z,\ g:=x^5-z^3 \in \Bbb{C}[x,y,z].$ The first thought is to prove that the quotient ring $\Bbb{C}[x,y,z]/\langle xy-z,x^5-z^3\rangle$ is an integral domain.

We observe that $f:=-z+xy\in \Bbb{C}[x,y][z]$ is irreducible in $\Bbb{C}[x,y][z]=\Bbb{C}[x,y,z]$ since $f$ has degree $1$ and $-1\in U(\Bbb{C}[x,y])=\Bbb{C^*}$ . And, from the general Eisenstein Criterion, we can take that $g$ is irreducible too.

1) Is it true that (since $f,g$ are irreducible ) $\gcd (f,g)=1\implies 1\in I \iff I=\Bbb{C}[x,y,z]?$ And if the answer is no, why?

2) I found in this post that although in $K[x]$ an irreducible polynomial generates a prime (maximal) ideal, now this is not always true, so we can not claim from irreducibility of $f,g$, that $I$ is prime.

Could you please give me a help?

Thank you in advance.

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No, this is not a prime. I'm going to take a different approach than the one you have begun on.

We see that $(xy -z, x^5 -z^3) = (xy-z, x^5 -x^3y^3) = (xy - z) + (x^3(x^2-y^3)$. From here, we can eliminate $z$ using $xy-z$: $$\frac{\mathbb{C}[x,y,z]}{(xy-z, x^5-z^3)} \cong \frac{\mathbb{C}[x,y]}{(x^3(x^2-y^3))}$$ This last ring is not an integral domain, as $x^3$ and $x^2-y^3$ are not zero in this ring, but their product is.

Hope this helps, let me know if not!

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  • $\begingroup$ Thank you very much for your answer. Could you please explain why $(xy -z, x^5 -z^3) = (xy-z, x^5 -x^3y^3)$? Why is this "legal"? And also could you explain the elimination you used? $\endgroup$ – Chris Jun 26 '18 at 21:39
  • $\begingroup$ And what will happen if we have the ideal $J:=(xyz-y^3,z^3-z^5)$? Could this elimination work somehow? $\endgroup$ – Chris Jun 26 '18 at 21:55
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    $\begingroup$ @Chris see if my answer (math.stackexchange.com/a/2833228/124095) helps to clear this up. $\endgroup$ – mweiss Jun 27 '18 at 3:37
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    $\begingroup$ @Chris yes, the answer mweiss has posted has the details fleshed out a bit more. $\endgroup$ – Brian Shin Jun 27 '18 at 12:17
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Here is a (possibly) simpler and more intuitive way to understand what happens when you take the quotient of $R=\mathbb C[x,y,z]$ by the ideal $I=\left< xy-z , x^5 - z^3 \right>$.

When we find a quotient of a ring by an ideal, we are essentially setting everything in the ideal equal to $0$. That amounts to introducing "replacement rules" that allow us to rewrite polynomials in $R$ in a different (hopefully simpler) form.

In this particular case, for example, the element $xy - z \in I$ tells us that in $R/I$, we have $xy-z=0$, so $xy=z$. That means that any polynomial in three variables can be rewritten as an equivalent polynomial in just two variables by replacing every occurrence of $z$ with $xy$. So, for example, the polynomials $x^2+y+z^3$ and $x^2+y+(xy)^3$ are different elements of $R$, but are identified in $R/I$. The practical upshot of this is that $\mathbb C[x,y,z]/\left<xy-z\right> \cong \mathbb C[x,y]$.

Now, what about the second generator of $I$, namely $x^5 - z^3$? Once we have already taken account of the replacement rule in the previous paragraph, this tells us that in $R/I$ the polynomial $x^5 - (xy)^3$ is set equal to $0$. Equivalently, we have $x^3(x^2-y^3) = 0$ as an identity in $R/I$.

But this clearly shows that $R/I$ is not an integral domain, because in $R/I$ we have $x^3 \ne 0$ and $x^2-y^3\ne 0$, but their product is $0$. If you want you can also express this in terms of the original ideal $I$: we have $x^3\notin I$ and $x^2-y^3\notin I$, but $x^3(x^2-y^3) \in I$. So $I$ is not prime.

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  • $\begingroup$ Thank you for your answer. So, "replacement rules" is a kind of technique to find the isomorphic ring. Right? $\endgroup$ – Chris Jun 27 '18 at 15:27
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    $\begingroup$ @Chris Right. I wouldn't use this as an argument in a formal proof -- there are all kinds of ways that generators of an ideal can have "hidden relationships" that are not obvious at first glance -- but as a heuristic for understanding the structure of the quotient ring I think it's extremely useful. And often it leads to the germ of an idea that can be the basis of a formal argument, as in the last two sentences of my answer. $\endgroup$ – mweiss Jun 27 '18 at 16:02
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  1. Why should $\gcd(f,g)\in I$? It looks like you're using that there exists $h,\ell$ so that $\gcd(f,g)=fh+g\ell$. But this fact, Bezouts identity, is true if the ring you are working in is a Bezout domain, i.e. a domain where the sum of two principal ideals is again principal. But $\Bbb C[x,y,z]$ is not such a domain: The ideal $(x)+(y) = (x,y)$ is not principal.
  2. Yup.

In general this kind of problem is not easy to do by hand. There is an algorithm for determining whether an ideal in $k[x_1,\dots, x_n]$ is prime, and it uses Grobner bases, where $k$ is any field. See, for example page 712 of Dummit and Foote's Abstract Algebra, Third Edition. This method allows you to do everything by hand, but can get a bit messy.

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  • $\begingroup$ Thank you for your anwer. Yes indeed, Bezout's identity was on my mind. I ll check Dummit and Foote's book for this. $\endgroup$ – Chris Jun 26 '18 at 21:45

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