It is well-known that a degree 2 or 3 polynomial over a field is reducible if and only if it has a root. But what about integral domains? Can we have a reducible polynomial over an integral domain having no roots in the domain?
Asked
Active
Viewed 1,388 times
3
-
"Can we have an irreducible polynomial over an integral domain having no roots in the domain?" do you mean reducible ? – Belgi Oct 21 '12 at 13:28
-
Yes, I meant reducible. – Reader Oct 21 '12 at 13:35
1 Answers
3
For example take $\mathbb{Z}$ and the polynomial $3(x^{2}+1)\in\mathbb{Z}[x]$
This polynomial have no roots in $\mathbb{Z}$ but $3\cdot(x^{2}+1)$ is a factorization
Belgi
- 23,150
-
-
1@Reader - why ? I recall that the definiton only requires that both polynimials are $\neq 0$ and that they are not invertible – Belgi Oct 21 '12 at 13:45
-
A polynomial $f (x) \in F [x]$ is called irreducible over $F$ if $deg(f) > 0$ and if its only factors are $c$ and $cf(x)$, where $c \in F$, $c\neq 0$, is any non-zero constant. – Reader Oct 21 '12 at 13:52
-
-
3@Reader: If you don't like the example, look at $4x^2-4x+1=(2x-1)(2x-1)$. Certainly reducible over $\mathbb{Z}$, but no integer zeros. – André Nicolas Oct 21 '12 at 18:52
-
-
Thanks. F is a field. What is definition of irreducible polynomial if F is not a field like, ring integral domain so on. – Reader Oct 21 '12 at 20:33
-
-
-
@Reader If you have found this answer helpful, I'd appriciate it if you could please accept by pressing on the v sign near the number 1 (left from my answer) – Belgi Oct 21 '12 at 21:26
-
-
-
Can somebody tell me what does it means to click on V? I am new at this forum. – Reader Oct 21 '12 at 21:48
-
@Reader - A. you can accept one answer (but you can un-accpt mine and choose anothr if you want) B. André posted a coment not an answer so you can not accept it unless he posts an answer (which I think he should!) – Belgi Oct 21 '12 at 22:14
-