Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A polynomial in a polynomial ring in one variable over a field generates a radical ideal iff it has no multiple roots. Is there a sufficient condition for a polynomial in several variables to generate a radical ideal? Like an ideal generated by a polynomial is prime if and only if it is irreducible.

share|cite|improve this question
Look up "square free" – Bill Cook Dec 18 '11 at 1:12
@Bill Cook: I know square free monomial ideals are radical and radical monomial ideals are square free. I was looking for conditions on polynomials that are not monomial. – Gene Simmons Dec 18 '11 at 1:36
@GeneSimmons, you should really look up square free! :D – Mariano Suárez-Alvarez Dec 18 '11 at 2:29
@MarianoSuárez-Alvarez: I tried various searches and looked up over a 100 articles but could not find anythIng close to the answer to my question. Perhaps I don't really understand the hint. I found some criteria for zero dimensional ideals in terms of square free polynomials, but not much else. If anyone has a pointed reference I would prefer that to random google searches. – Gene Simmons Dec 18 '11 at 4:20
@Gene: Prove it! If $f$ is squarefree and $f|g^n$, then for every irreducible factor $p$ of $f$, $p|g^n$, hence $p|g$. Therefore (since distinct irreducible factors are relatively prime), the product of all distinct irreducible factors of $f$ divides $g$; but this product is (an associate of) $f$, because $f$ is squarefree. So if $f$ is squarefree, $g^n\in (f)\Rightarrow g\in (f)$, so $(f)$ is radical. Conversely, if $f$ is not squarefree, then the squarefree root of $f$ has a power that lies in $(f)$ but does not itself lie in $f$. – Arturo Magidin Dec 18 '11 at 5:09
up vote 2 down vote accepted

Let $k$ be a field and consider the polynomial ring $A = k[x_1,...,x_n]$.

Claim: Given $f \in A - \{0\}$, (f) is radical if and only if $f$ factors into a product of irreducibles of multiplicity $1$.


$\Leftarrow$: We know $A$ is a UFD. So, let $f = f_1...f_m$ be a product of $f$ into irreducible factors such that for all $i \neq j$, $(f_i) \neq (f_j)$. Then $(f) = (f_1...f_m) = (f_1) \cap ... \cap (f_m)$ (I am using unique factorization for the second equality). Thus, $(f)$ is an intersection of prime ideals of $A$ and hence radical.

$\Rightarrow$: Suppose $(f)$ is radical. Again, let $f = {f_1}^{e_1}...{f_m}^{e_m}$ be a product of $f$ into irreducibles where $i \neq j$ $\Rightarrow$ $(f_i) \neq (f_j)$.

Our goal is to show that each $e_i = 1$. Well, suppose not. Then there exists $e_i$ such that $e_i > 1$. Then $({f_1}^{e_1}...{f_i}^1...{f_m}^{e_m}) \subset (f) \subset ({f_1}^{e_1}...{f_i}^1...{f_m}^{e_m})$. The first inclusion is because ${f_1}^{e_1}...{f_i}^1...{f_m}^{e_m} \in Rad((f)) = (f)$, and the second inclusion follows from the fact that ${f_1}^{e_1}...{f_i}^1...{f_m}^{e_m}|f$.

But, this means that there is some $u \in A^*$ such that ${f_1}^{e_1}...{f_i}^1...{f_m}^{e_m} = u{f_1}^{e_1}...{f_i}^{e_i}...{f_m}^{e_m}$, which contradicts unique factorization.

share|cite|improve this answer
Thanks. One question, why is the ideal generated by the product of the irreducible factors equal to the intersection of the ideals generated by the individual factors? – Gene Simmons Dec 18 '11 at 5:20
$\subset$ is elementary, and $\supset$ follows from unique factorization. – Rankeya Dec 18 '11 at 5:22
Thanks. One follow up question. This doesn't extend to non-principal ideals right? – Gene Simmons Dec 18 '11 at 5:23
What is the precise statement you are trying to make for non-principal ideals? – Rankeya Dec 18 '11 at 5:27
Are ideals generated by several square free polynomials radical? – Gene Simmons Dec 18 '11 at 5:33

Let $(p_1),\dots,(p_n)$ be distinct prime ideals of a unique factorization domain, and let $k_1,\dots,k_n$ be positive integers. Then the radical of $$ (p_1^{k_1}\cdots p_n^{k_n}) $$ is clearly $$ (p_1\cdots p_n). $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.