When do polynomials have common roots? In my workbook is given such an exercise and, so , can you write please what's the condition for this thing to happen, so that two polynomials have one or more common roots. Thank you!
[Math] When do polynomials have common roots
polynomials
Related Solutions
Your proof seems to work, but it looks incomplete because you need to use more explicitly the fact that the ring of polynomials is an UFD.
The condition that the resultant $\mathrm{Res}(f,g)=0$ means that there are some polynomials $p,q$ such that $\deg p<\deg f$, $\deg q<\deg g$ and $fq=gp$ (so $p/q=f/g$). In a UFD this is equivalent to having a common factor of $f$ and $g$.
To complete your approach you must replace the coefficient ring by its quotient field. You need the unique factorization at the end when you simplify the coefficients to be ring elements again.
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For zero divisor free rings, without unique factorization, it will not work. For example, consider the polynomials $x^2-4y^2$ and $x^2+4xy+4y^2$. In some rings, like $(\mathbb{R}[y])[x]$, they have the common factor $x+2y$. In other rings, for example in $(\mathbb{Z}[4y,4y^2])[x]$, they don't. However, the resultants will be the same zero polynomial in all rings. So this method does not work if the coefficients are from the ring $\mathbb{Z}[4y,4y^2]$.
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Another possible approach is building a single (but huge) resultant. If $f,g\in\mathbb{R}[x_1,\dots,x_n]$ are nonzero polynomials then take all polynomials of the form $x_1^{a_1}\cdots x_n^{a_n}f$ with $a_1+\cdots+a_n<\deg g$ and $x_1^{b_1}\cdots x_n^{b_n}g$ with $b_1+\cdots+b_n<\deg f$. These polynomials are linearly dependent if and only if there are some polynomials $p,q$ with $\deg p<\deg f$, $\deg q<\deg g$ and $p/q=f/g$.
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Sorry, I cannot recommend you books other than I can find by internet search... :-(
If there is a root in common, then using the Euclidean algorithm for polynomials to find the highest common factor either (a) gives a lower degree polynomial of which the root is a factor (so one of the polynomials at least is not minimal); or (b) tells you that the polynomials are the same.
(This makes assumptions about the admissible coefficients of the polynomials - the Euclidean algorithm has to apply, but this will almost certainly be the case in the circumstances you envisage)
Best Answer
Two polynomials have a common root if and only if their resultant vanishes.
The resultant $R(p(x), q(x))$ of two polynomials of degrees $m$ and $n$, respectively, is the determinant of the $(m + n) \times (m + n)$ matrix defined as follows. Write the coefficient of $p(x)$ in the first row followed by $n - 1$ zeros. In the next row the coefficients are displaced one place to the right, with one zero to the left and $n - 2$ zeros to the right. Continue in this fashion until the $nth$ row is $n - 1$ zeros followed by the coefficients of $p(x)$. For the last $m$ rows, we do the same thing with $p(x)$ and $q(x)$ interchanged.
With your example, we have the resultant as: $$ R(p(x), q(x)) = \mbox{det}\left(\begin{array}{cccccccc} 22 & 33 & -16a & -3 & 2 & 0 & 0 & 0\\ 0 & 22 & 33 & -16a & -3 & 2 & 0 & 0\\ 0 & 0 & 22 & 33 & -16a & -3 & 2 & 0\\ 0 & 0 & 0 & 22 & 33 & -16a & -3 & 2\\ 11 & 33 & 21 & -2a & -2 & 0 & 0 & 0\\ 0 & 11 & 33 & 21 & -2a & -2 & 0 & 0\\ 0 & 0 & 11 & 33 & 21 & -2a & -2 & 0 \\ 0 & 0 & 0 & 11 & 33 & 21 & -2a & -2 \end{array}\right).$$
Computer solution gives $a = \dfrac{3}{16}, \dfrac{297}{128},$ or $\dfrac{3}{2}$.