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9.11 Relatively prime polynomials

Let $K$ be an algebraically closed field. Then the ring $K[x]$ has a very simple ideal structure as we saw in Lemma 9.10.2. In particular, every polynomial $P \in K[x]$ can be written as

\[ P = c(x - \alpha _1) \ldots (x - \alpha _ n), \]

where $c$ is the constant term and the $\alpha _1, \ldots , \alpha _ n \in k$ are the roots of $P$ (counted with multiplicity). Clearly, the only irreducible polynomials in $K[x]$ are the linear polynomials $c(x - \alpha )$, $c, \alpha \in K$ (and $c \neq 0$).

Definition 9.11.1. If $k$ is any field, we say that two polynomials in $k[x]$ are relatively prime if they generate the unit ideal in $k[x]$.

Continuing the discussion above, if $K$ is an algebraically closed field, two polynomials in $K[x]$ are relatively prime if and only if they have no common roots. This follows because the maximal ideals of $K[x]$ are of the form $(x - \alpha )$, $\alpha \in K$. So if $F, G \in K[x]$ have no common root, then $(F, G)$ cannot be contained in any $(x - \alpha )$ (as then they would have a common root at $\alpha $).

If $k$ is not algebraically closed, then this still gives information about when two polynomials in $k[x]$ generate the unit ideal.

Lemma 9.11.2. Two polynomials in $k[x]$ are relatively prime precisely when they have no common roots in an algebraic closure $\overline{k}$ of $k$.

Proof. The claim is that any two polynomials $P, Q$ generate $(1)$ in $k[x]$ if and only if they generate $(1)$ in $\overline{k}[x]$. This is a piece of linear algebra: a system of linear equations with coefficients in $k$ has a solution if and only if it has a solution in any extension of $k$. Consequently, we can reduce to the case of an algebraically closed field, in which case the result is clear from what we have already proved. $\square$

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