# When does a polynomial generate a radical ideal?

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.

• 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-Álvarez 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

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

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

Proof:

$$\Leftarrow$$: We know $$A$$ is a UFD. So, let $$f = f_1\cdots f_m$$ be a product of $$f$$ into irreducible factors such that for all $$i \neq j$$, $$(f_i) \neq (f_j)$$. Then $$(f) = (f_1\cdots f_m) = (f_1) \cap \cdots \cap (f_m)$$ (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 $$f = {f_1}^{e_1}\cdots {f_m}^{e_m}$$ 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 $$({f_1}^{e_1}\cdots {f_i}^1\cdots {f_m}^{e_m}) \subseteq (f) \subseteq ({f_1}^{e_1}\cdots {f_i}^1\cdots {f_m}^{e_m})$$. The first inclusion is because $${f_1}^{e_1}\cdots {f_i}^1\cdots {f_m}^{e_m} \in \mathrm{Rad}((f)) = (f)$$, and the second inclusion follows from the fact that $${f_1}^{e_1}\cdots {f_i}^1\cdots {f_m}^{e_m}\mid f$$.

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

• 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 $$(p_1\cdots p_n).$$