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

Let us fix a field $K$, and let us consider the principal ideal ring $K[x]$. If I is an ideal, since $K[x]$ is a PID, we can write $I = (p(x))$ for some polynomial $p(x)$. Now, let us say that a set of generators $S= \{x_1, \ldots , x_n\}$ for I is irredundant if no subset of S of cardinality less than n generates I.

Can we construct, for every $n \geq 1$, and any non-zero ideal an irredundant set of generators of cardinality $n$? For example, for $n= 2$, the ideal $(x)$ has an irredundant set of generators $\{x+x^2,x^2\}$. I have tried doing some induction, but it didn't amount to much, and neither did explicitly trying to construct an irredundant set for an arbitrary non-zero ideal. So, does anyone see how to do it, and if so, any hints?

share|cite|improve this question
up vote 3 down vote accepted

In general, a set of elements generates $\left<r\right>$ in a PID if, by definition, their GCD is $r$.

Let $r(x)\in K[x]$. Let $p_1(x),p_2(x),\dots,p_n(x)$ be $n$ distinct prime monic polynomials. Let $$a_i(x)=r(x)\frac{p_1(x)\dots p_n(x)}{p_i(x)}$$

Then the gcd of $\{a_i(x)\}$ is $r(x)$, but the GCD of any proper subset is not.

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.