# Number of generators of ideals in a PID.

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?

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## 1 Answer

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.

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