Timeline for Is a polynomial with 1 very large coefficient irreducible?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Dec 14, 2009 at 14:07 | comment | added | j.p. | @Gjergji: Looking at $fg \bmod 2$ rsp. $fg \bmod 3$ you see that () $f$ and $g$ have exactly one coefficient NOT divisible by $2$ rsp. $3$. If $f$ is monic then its constant coefficient is divisible by $6$. Hence the constant coefficient of $g$ is $\pm 1$. Because of () all other coefficients of $f$ and $g$ are divisible by $6$. Thus $fg \bmod 6$ has exactly one nonzero coefficient contradicting the definition of the $a_i$'s. (If the leading coefficient of $f$ is divisible by $6$, then leading coefficient of $g$ is $\pm 1$. ...) | |
| Dec 12, 2009 at 23:33 | comment | added | Gjergji Zaimi | @jp: why can't any of f and g be monic? | |
| Dec 11, 2009 at 19:38 | vote | accept | Gjergji Zaimi | ||
| Dec 10, 2009 at 14:45 | comment | added | j.p. | Back to Vladimir's suggestion with (4, 9, 6, 6, ...). If $0 < k < n$ then $f$ rsp. $g$ have leading coefficient $2$ and constant coefficient $3$ rsp. $3$ and $2$ and all other coefficients are multiples of $6$. If I didn't make a stupid mistake, then considering $fg \bmod 36$ leads to a contradiction (at least for $1 < k < n-1$). | |
| Dec 9, 2009 at 18:15 | comment | added | Vladimir Dotsenko | okay, I don't know what happened to the long google books url in the previous comment, so let me re-type it in a better way: tinyurl.com/prasolov | |
| Dec 9, 2009 at 18:12 | comment | added | Vladimir Dotsenko | @jp: sorry for not providing a reference - I was in a hurry. An excellent source on many a theorem about polynomials is <a href="books.google.com/… book</a> (for facts I am referring to, see page 53, Dumas' theorem (and a bit before this theorem), but there are lots of other useful things there). | |
| Dec 9, 2009 at 18:11 | comment | added | Vladimir Dotsenko | @Gjergji: good point - I was thinking along the same line after I walked in the street having written that suggestion. However, what you say about how to mend my example seems to be a good idea. | |
| Dec 9, 2009 at 16:19 | comment | added | j.p. | If someone needs a reference for the statement about the Newton polygon (like I did), take a look at the corollary at the bottom of page two in math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf | |
| Dec 9, 2009 at 16:08 | comment | added | Gjergji Zaimi | Thanks! The suggestion works (I think) with (4,9,25,30,30...,30). The problem with (4,9,6...) is when a_k=4,a_{n-k}=9 for some k, which by looking at the Newton polygon mod2, mod3 gives the polynomial as a product fg where f is degree k and f,g are irreducible mod2, mod3. I couldn't find a contradiction in that case. When we introduce 25 such "symmetric" factorizations are ruled out. I haven't checked when 4,9 or 25 are leading yet (Eisenstein doesn't apply anymore). | |
| Dec 9, 2009 at 14:01 | history | answered | Vladimir Dotsenko | CC BY-SA 2.5 |