# Methods to check if an ideal of a polynomial ring is prime or at least radical

I am looking for methods to check whether a given ideal in $K[x_0,\dots,x_n]$ is prime. I mean something you can effectively use in some concrete non-trivial example. To be more explicit, I am working in $K[x_0,\dots,x_3]$ ($K$ algebraically closed) and I have got these four polynomials:

$f_1=x_0x_3-x_1x_2,$

$f_2=x_0^2x_2-x_1^3,$

$f_3=x_1x_3^2-x_2^3,$

$f_4=x_0x_2^2-x_1^2x_3$.

Let then $J=(f_1,f_2,f_3)$ and $I=(f_1,f_2,f_3,f_4)$. I would like to prove that $I=\operatorname{rad}(J)$. One can easily check that $f_4^2 \in J$, so we have $I\subset \operatorname{rad}(J)$. On the other hand we have $J\subset I\subset \operatorname{rad}(J)\subset \operatorname{rad}(I)$, and so $\operatorname{rad}(I)=\operatorname{rad}(J)$. And here is my problem: it would suffice to show that $I$ is prime or at least radical to have what we want, but I find $I$ complicated enough to check this directly frome the definition and first properties. So I am asking if anyone knows some method which can easily be applied to this ideal (which I know to be prime thanks to a computer algebra system..but I need a proof). I notice that $I$ is a homogeneous ideal, maybe this can help, I don't know if there is any ad hoc method for homogeneous ideals.

Edit: Explicit calculations now included at the bottom. Thanks go to Dylan Moreland for suggesting I use Sage to compute Groebner bases.

Edit 2: I have now included a lemma with proof towards the bottom that extends this method so that it can show that an ideal is primary, even if it is not prime.

Using any techniques I'm familiar with, proving an ideal prime (radical) is generally painful. The method I use below is tailored to your specific example but can be adapted to work more generally. If we want to show that $I$ is prime (radical), the easiest way IMHO is to show that $K[x_0,\ldots,x_3]/I$ is an integral domain (has only nilpotent zero-divisors). The methods that follow work to prove an ideal prime, but not to prove an ideal radical.

Note: I abuse notation, using $x_0,x_1$, etc. for both elements of $K[x_0,\ldots,x_3]$ and their images under various homomorphisms. I do not believe this will result in any confusion, but if it does I apologize and will attempt to clarify.

Outline of Method: We begin by determining that certain elements are not zero-divisors, hence localizing to invert these elements does not change whether or not the ring is an integral domain. Then we use substitution to reduce the number of generators and eventually get to just $1$ homogeneous polynomial, which we show is irreducible by elementary methods.

The key is judicious localization. First note that the image of $x_1$ in $R$ is not a zero-divisor. To see this, suppose $x_1f\in I$ for some $f\in K[x_0,\ldots,x_3]$, so we have $f\in (I : (x_1))$. We want to show that $f\in I$, or equivalently that $(I:(x_1)) = I$. This can be done using Groebner bases. Since $K[x_0,\ldots,x_3]$ is an integral domain, $$(I : (x_1)) = \frac{1}{x_1}(I\cap (x_1)) = \frac{1}{x_1}\left((tI+(1-t)(x_1))\cap K[x_0,\ldots,x_3]\right)$$ which is generated by the elements of a Groebner basis for $tI+(1-t)(x_1)$ that are independent of $t$ divided by $x_1$, and in fact this basis is a Groebner basis. Once you have this, you can compute a Groebner basis for $I$ and verify that the two ideals are the same by performing polynomial division on each generator of each ideal by all the generators of the other ideal, which should return $0$. It so happens that my desktop, which has mathematica and a fast processor, is a thousand miles away at the moment, so I will not do these calculations at this time (you however are free to, and I may do so later).

The following requires a couple of theorems concerning localization and polynomial rings. Having established that $x_1$ is not a zero-divisor, we let $S_1 = \{1,x_1,x_1^2,\ldots\}$ and note that $R$ is an integral domain iff $R_{S_1}$ is. We want to show that $x_2$ is not a zero divisor in order to eliminate another variable. We can certainly use the same technique as before to verify that it is not a zero-divisor in $R$, and if $fx_2=0$ in $R_{S_1}$ then we can clear denominators by multiplying by $x_1^k$ to get $(fx_1^k)x_2=0$ in $R$ and $fx_1^k=0$ iff $f=0$, so this verifies that it is not a zero-divisor in $R_{S_1}$. Again, I will not do these calculations here but will assume that they check out. Since $x_2$ is not a zero-divisor in $R_{S_1}$, we let $S_2 = \{1,x_2,x_2^2,\ldots\}$ and note that $R_{S_1}$ is an integral domain iff $R' = R_{S_1S_2}\cong (R_{S_1})_{S_2}$ is. Because $x_1,x_2$ are invertible in $R'$, we can let $x_0' = x_0+x_2^{-2}x_1^2x_3$ to get $$R' \cong K[x_0,x_1,x_2,x_3]_{S_1S_2}/I_{S_1S_2} = (K[x_1,x_2])_{S_1S_2}[x_0',x_3]/I_{S_1S_2}$$ and using this we eliminate variables mercilessly $$\begin{eqnarray*}I_{S_1S_2} &=& ((x_0'-x_2^{-2}x_1^2x_3)x_3-x_1x_2, (x_0'-x_2^{-2}x_1^2x_3)^2x_2-x_1^3, x_1x_3^2-x_2^3, (x_0'-x_2^{-2}x_1^2x_3)x_2^2-x_1^2x_3)\\ &=& ((x_0'-x_2^{-2}x_1^2x_3)x_3-x_1x_2, (x_0'-x_2^{-2}x_1^2x_3)^2x_2-x_1^3, x_1x_3^2-x_2^3, x_0'x_2^2)\\ &=& (x_2^{-2}x_1^2x_3^2-x_1x_2, x_2^{-3}x_1^4x_3^2-x_1^3, x_1x_3^2-x_2^3, x_0')\\ &=& (x_1x_3^2-x_2^3, x_1x_3^2-x_2^3, x_1x_3^2-x_2^3, x_0')\\ &=& (x_1x_3^2-x_2^3,x_0')\end{eqnarray*}$$ giving us a much nicer set of generators to work with. From this we see $$R' \cong (K[x_1,x_2])_{S_1S_2}[x_3]/(x_1x_3^2-x_2^3)\cong \left(K[x_1,x_2,x_3]/(x_1x_3^2-x_2^3)\right)_{S_1S_2}$$ which is an integral domain iff $K[x_1,x_2,x_3]/(x_1x_3^2-x_2^3)$ is, so we want to show that $x_1x_3^2-x_2^3$ is prime in $K[x_1,x_2,x_3]$. Since $K[x_1,x_2,x_3]$ is a UFD, we need only show that $x_1x_3^2-x_2^3$ is irreducible. There are algorithms for this in general, but we can simplify this case considerably by observing that $x_1x_3^2-x_2^3$ is homogeneous. If it is not irreducible, then it can be factored into two homogeneous polynomials of degree $1$ and $2$. This can be easily checked, as if $$\begin{eqnarray*}x_1x_3^2-x_2^3&=&(a_1x_1+a_2x_2+a_3x_3)(b_1x_1^2+b_2x_1x_2+b_3x_1x_3+b_4x_2^2+b_5x_2x_3+b_6x_3^2)\\ &=&(a_1b_6+a_3b_3)x_1x_3^2+a_2b_4x^3+a_2b_3x_1x_2x_3+a_2b_6x_2x_3^2+\text{ other junk}\end{eqnarray*}$$ which implies $a_2b_3=a_2b_6=0$ and $a_1b_6+a_3b_3\neq 0,a_2b_4\neq 0$, which is easily seen to be a contradiction. Hence $R'$ is an integral domain, so $I$ is prime.

Note on Radical Ideals: Proving an ideal radical is, in my experience, generally more difficult. I try to figure out the primary decomposition of the ideal, and then show that each primary ideal in the decomposition is in fact prime by something like the above method. However, if it is true that inverting a non-zero-divisor does not change whether or not a ring has only nilpotent zero-divisors, then maybe the method above can be modified to prove that ideals are radical.

Lemma: If $x\in R$ is not a zero-divisor and we let $S = \{1,x,x^2,\ldots\}$, then $R$ has only nilpotent zero-divisors iff $R_S$ does.

Proof: Suppose $R$ has some non-nilpotent $y,z$ such that $yz=0$. Since $x$ is not a zero-divisor, the homomorphism $R\to R_S$ is injective so $\frac{y^n}{1},\frac{z^n}{1}\neq 0$ for any $n$ so $\frac{y}{1},\frac{z}{1}$ are not nilpotent, yet $\frac{y}{1}\frac{z}{1}=\frac{0}{1}$ so $R_S$ has only non-nilpotent zero-divisors. Suppose now that $R_S$ has some non-nilpotent $\frac{y}{1},\frac{z}{1}$ such that $\frac{y}{1}\frac{z}{1}=\frac{0}{1}$. Clearly $y,z$ must be non-nilpotent, but $yz=0$ so $R$ has non-nilpotent zero-divisors.

Computations for $(I : (x_1))$: I forgot to mention that the Groebner basis must be computed with respect to a lexicographical ordering which places $t$ first. According to Sage, the following is a Groebner basis for $tI+(1-t)(x_1)$: $$\{x_1^4 - x_0^2x_1x_2, tx_0^2x_2 - x_1^3, tx_0x_2^2 - x_1^2x_3, x_0x_1x_2^2 -x_1^3x_3, tx_2^3 - x_1x_3^2,$$ $$x_1x_2^3 - x_1^2x_3^2, x_1^2x_2 - x_0x_1x_3, tx_0x_3 - x_1x_2, tx_1 - x_1\}$$ which after eliminating the terms involving $t$ and dividing by $x_1$ gives us $$(I : (x_1))=(x_1^3 - x_0^2x_2, x_0x_2^2 -x_1^2x_3, x_2^3 - x_1x_3^2, x_1x_2 - x_0x_3)$$ which is equal to $I$ because $$\begin{eqnarray*} x_1^3 - x_0^2x_2 &=& -(x_0^2x_2-x_1^3)\\ x_0x_2^2-x_1^2x_3 &=& x_0x_2^2-x_1^2x_3\\ x_2^3 - x_1x_3^2 &=& -(x_1x_3^2-x_2^3)\\ x_1x_2 - x_0x_3 &=& x_1x_2 - x_0x_3 \end{eqnarray*}$$ hence $x_1$ is not a zero-divisor in $R$.

Computations for $(I : (x_2))$: Sage gives the following Groebner basis for $tI+(1-t)(x_2)$: $$\{tx_1^3 - x_0^2x_2, x_1^3x_2 - x_0^2x_2^2, x_0x_2^3 - x_1^2x_2x_3, x_2^4 -x_1x_2x_3^2, tx_1^2x_3 - x_0x_2^2,$$ $$tx_1x_3^2 - x_2^3, x_1x_2^2 - x_0x_2x_3,tx_0x_3 - x_1x_2, tx_2 - x_2\}$$ which after eliminating the terms involving $t$ and dividing by $x_2$ gives us $$(I : (x_2))=(x_1^3 - x_0^2x_2, x_0x_2^2 - x_1^2x_3, x_2^3 -x_1x_3^2, x_1x_2 - x_0x_3)$$ which is identical to $(I : (x_1))=I$ hence $x_2$ is not a zero-divisor in $R$, which means it is not a zero-divisor in $R_{S_1}$.

By using the same technique I've used here one can prove that $K[x_0,\dots,x_3]/I$ is isomorphic to $K[s^4,s^3t,st^3,t^4]$. In particular, this shows that $I$ is a prime ideal.

• I've looked at your link and wow that's smart! I'm now trying to adapt that solution to my ideal, but I'm having trouble doing the "division with reminder" step. I get a remainder which a priori is a combination of a lot of monomials of degree 3 (whereas in your link you find just 3 monomials of degree 2)....am I missing something or things are just more complicated because my generators have degree 3? – Lor Jan 31 '13 at 14:16
• @Lor If I remember well, there are essentially 7-8 types of monomials (not of degree 3, but containing 2 or 3 variables; be careful, one type of these monomials is surely in two variables, $x_2$ and $x_3$) that can appear in the remainder and can't cancel any of them when replace $x_0$ by $s^4$, and so on. – user26857 Jan 31 '13 at 16:37
• ok I'll go through it, thank you for your help. Just two more questions, if you don't mind. 1) how do I know a priori the number of parameters ( I mean s,t,...) and how do I find the mapping $x_0 \to s^4, \dots..$ ? it looks a lot god-given to me 2) this technique is quite the same as giving a parametrization in $s$ and $t$ of my variety $K[\dots]/I$, isn't it? – Lor Jan 31 '13 at 17:04
• 1) I'm not a good enough geometer to answer this question. I've posted a similar question here, but so far no one posted an answer. 2) I think so. – user26857 Jan 31 '13 at 17:26
• I'll keep checking that page then. Thanks a lot again! – Lor Jan 31 '13 at 17:42