# Sum of polynomials with no common factors

I have come across this problem in a set of exercises leading to a proof of the Jordan Normal Form. It begins with taking a polynomial $h(x)$ such that $h(L)\equiv 0$ for a linear operator $L$, and factoring it as $(x-\lambda_1)^{n_1}\dotsm(x-\lambda_k)^{n_k}$. Then define the polynomial $g_i(x)=\prod_{j\neq i}(x-\lambda_j)^{n_j}$.

The exercise: Given the $k$ polynomials $g_i(x)$, show that there exist polynomials $f_i(x)$ such that $\sum_i f_i(x)g_i(x)=1$. (There is a hint: what is the greatest common factor of the $g_i$'s?)

-
What are $\lambda_i$, i.e. what field are you working with? Is it $\mathbb R$? – awllower Jul 28 '13 at 14:26
I think $\mathbb{C}$, so that the factorisation of $h$ as linear factors is always possible. – Mark Jul 28 '13 at 14:36
OK. Thanks for the clarification. – awllower Jul 28 '13 at 14:51

## 1 Answer

Let $k$ be a field.

Notice that the $g_i$'s have no factor in common, and that $k[x]$ is a PID. If $R$ is a PID, and therefore a UFD, and $a_1,\ldots,a_k\in R$ share no common factor in $R$, then there exist $b_1,\ldots,b_k\in R$ such that $\sum a_ib_i=1$. This is exactly the result you need.

In general, if $d=(a_1,\ldots,a_k)$, then $d$ is a linear combination of the $a_i$.

An explicit argument, still using the fact that $k[x]$ is a PID. Since $k[x]$ is a PID, we know there is some $p(x)$ such that $(g_1(x),\ldots,g_k(x))=(p(x))$. This means that $p(x)$ divides each $g_i(x)$. Since the $g_i(x)$ are relatively prime, $p(x)$ must be a unit, so $(g_1(x),\ldots,g_k(x))=(p(x))=k[x]$. By definition of an ideal generated by elements $g_i(x)$, there must be $f_i(x)$ such that $\sum f_i(x)g_i(x)=1$.

-
Would you be able to explain this to me in a less abstract way? I'm not very knowledgable on abstract algebra! – Mark Jul 28 '13 at 14:46
@MarkC: See if what I added is helpful. – Jared Jul 28 '13 at 20:19