Necessary and sufficient condition for two polynomials to generate coprime ideals

abstract-algebraidealspolynomialsring-theory

This question concerns the ring of polynomials with integer coefficients, $R=\mathbb Z[X]$. $R$ is not a principal ideal domain, and two elements $p,q\in R$ that are coprime does not necessarily generate coprime ideals, i.e. it is possible to have $\langle p \rangle+\langle q \rangle\neq R$.

I am investigating whether there is a simple way to decide whether two coprime elements $p,q$ generate coprime ideals. For example, the $\langle 2x+3 \rangle+\langle 4 \rangle\neq R$.

I have tried a few examples, but it turns out that every time I am directly expanding the coefficients:
$$
a(x)p(x)+b(x)q(x)=\sum_{r=0}^\infty \left(\sum_{i+j=r} (a_ip_j+b_iq_j) \right) x^r=1.
$$
So we may deduce that $a_0p_0+b_0 q_0=1$, so $p_0, q_0$ are coprime. And then the four integers $p_1, q_1, p_0, q_0 $ are $\mathbb Z$-linearly dependent, and so on. This is essentially a system of linear equations of infinitely many unknowns, and I don't know how to tackle the inverse of an infinite matrix – after all, it is NOT a complex matrix, but an integer matrix.

I am not quite getting the essence of the issue here. Are there any simple ways to decide whether of not $\langle p \rangle+\langle q \rangle= R$? Is there an algorithm for it?

In general, how to judge if two elements generate coprime ideals in a ring?

Best Answer

If the principal ideals generated by $p,q$ are coprime, then their highest degree coefficients are coprime, since if $g$ is the greatest common divisor of the highest degree coefficients of $p,q$ then $g$ also divides the highest degree coefficient of every element of $(p)+(q)$. Suppose $\deg p \le \deg q=n$.

Lemma. The highest degree coefficients of $p,q$ are coprime if and only if there is a monic polynomial of the form $ap+bq$ with degree at most $n$.

Proof. Multiplying $p$ by $x^k$ for some $k$, we can make the degree of $x^kp$ equal to the degree of $q$. Then the degree $n$ coefficients of $x^kp,q$ are coprime iff there is an integer linear combination making them 1, iff there is an integer linear combination of $x^kp,q$ that is monic. $\blacksquare$

Here is an algorithm:

  1. Check that the highest degree coefficients of $p,q$ are coprime. If not, $(p)+(q)\ne (1)$, so stop.
  2. Otherwise, there is a monic polynomial of degree $n$ of the form $ap+bq$, divide $p$ by $ap+bq$ and replace $p$ by the remainder.
  3. Repeat.

This produces a sequence of generators of the ideal $(p)+(q)$ of decreasing degree. Either it terminates in some round of step 1 showing that the polynomials do not generate coprime ideals, or we get a monic polynomial of degree 0 in $(p)+(q)$, i.e. $(p)+(q)=(1)$.

In your example, $2x+3,4$ do not have highest degree coefficients coprime, so $(2x+3)+(4)\ne (1)$ as you said.

Related Question