Given three polynomials $a(x), b(x)$, and $c(x)$ satisfying $\gcd(a(x), b(x), c(x))=1$

Question

Given three nonzero polynomials $a(x), b(x)$ and $c(x)$ satisfying $\gcd(a(x), b(x), c(x))=1$.

Please help me prove that there exists six polynomials $f(x), g(x), h(x), u(x), v(x), w(x)$ such that

$$\begin{vmatrix} a(x) & b(x) & c(x) \\ f(x) & g(x) & h(x) \\ u(x) & v(x) & w(x) \end{vmatrix}=1$$

What does $(a(x),b(x),c(x)) = 1$ stand for? Should you equations hold for all $x$ (in $\mathbb{R}$), or just for one? Where are your own ideas? — Dirk, May 11 '17 at 10:05
@Bemte It means the greatest common divisor of three polynomials.I have edited in the problem just now. I am so sorry. — user444628, May 11 '17 at 10:09
They are elements of a principal ideal domain. That is precisely the context of the other question. Clear duplicate imho. — MooS, May 11 '17 at 12:24

score 0 · Accepted Answer · edited May 11 '17 at 10:30

0

Let $G(x) := \gcd(a(x),b(x))$.

Then there are polynomials $A(x), B(x)$ such that $a(x)A(x) + b(x)B(X) = G(x)$.

Now we know that $\gcd(G(x),c(x)) = 1$.

Thus, there are polynomials $C(x),D(x)$ such that $G(x)D(x) + c(x)C(x) = 1$.

Expand this to read $$a(x)A(x)D(x) + b(x)B(X)D(x) + c(x)C(x) = 1$$

Look at the formula for the determinant of a $3 \times 3$ matrix and try to take it from there.

edited May 11 '17 at 10:30

lioness99a

answered May 11 '17 at 10:14

Dirk

Thank you but I have come to this. I suspect that the difficulty is in making 6 polynomials satisfying the equation. – user444628 May 11 '17 at 10:21
@MooS has got the answer. – user444628 May 11 '17 at 13:01

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