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In B. L. van der Waerden's Algebra it's said that one can considerably simplify usage of Kronecker's method for polynomials over the ring of integers by factoring the given polynomial modulo 2 and possibly modulo 3, so that one gets an idea what degrees the possible factor polynomials might have, and to what residue classes the coefficients modulo 2 and 3 might belong.

I don't have a clue how that information might help. Can someone explain that? I would be really grateful for a little example too, that would help me for good.

Moreover, I might be wrong in understanding what is a polynomial modulo ring element. Is that just polynomial with coefficients modulo that element?

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Suppose that $p(x)$, with integer or rational coefficients, factors over the rationals into $p_1(x)p_2(x)...p_k(x)$. Then, modulo $q$, it must factor in at least $k$ factors (not necessarily of the same degree).

So, the idea is that factorizations modulo $q$ gives you a bound from below on the number of irreducible factors over the rationals. In the extreme case that you find that modulo $q$ the polynomial is irreducible of the same degree, then the polynomial is forced to be irreducible over the rationals.

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  • $\begingroup$ Beware that the claim holds only for monic polynomials, e.g. $\, (30x+1)^n\,$ has $n$ prime factors in $\,\Bbb Z[x]\,$ but none $\bmod 2,3,5\ \ \ $ $\endgroup$ – Gone Feb 3 at 18:17
  • $\begingroup$ @BillDubuque $0$ is a lower bound of $n$. For the implication about irreducibly one doesn't need to restrict to monic polynomials. It is enough to remember the degree of the original polynomial. $\endgroup$ – Blue Feb 3 at 18:44
  • $\begingroup$ Your claim implies $\,(30x+1)^n\,$ has at least $n$ factors $\!\bmod q$. That is false for $\,n\ge 1\,$ and $\,q = 2,3,5.\ \ $ $\endgroup$ – Gone Feb 3 at 19:13
  • $\begingroup$ @BillDubuque There is no claim as to the nature of those factors. $(30x+1)^n$ modulo $2$ is $1\cdot 1\cdot ...\cdot 1$. One only needs to keep track of the degrees. $\endgroup$ – Blue Feb 3 at 19:23
  • $\begingroup$ Glad to see you edited it. Note also that a factor might even vanish $(\equiv 0)$ if the polynomial is not primitive. $\endgroup$ – Gone Feb 3 at 19:41

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