# Irreducible Polynomial 2

• Jun 8th 2009, 07:59 AM
osodud
Irreducible Polynomial 2
Show that a polynomial of order 2 or 3 is irreducible on a field $F$ if has a root in $F$.

Use this to show that the polynomial $x^3-3x=1$ is irreducible in $Z[x]$

Eisenstein's creiterio ??
• Jun 8th 2009, 08:54 PM
NonCommAlg
Quote:

Originally Posted by osodud
Show that a polynomial of order 2 or 3 is irreducible on a field $F$ if has a root in $F$.

Use this to show that the polynomial $x^3-3x=1$ is irreducible in $Z[x]$

Eisenstein's creiterio ??

$x^3-3x=1$ is not a polynomial. it's an equation! did you mean $x^3-3x-1$?
• Jun 8th 2009, 08:57 PM
osodud
Quote:

Originally Posted by NonCommAlg
$x^3-3x=1$ is not a polynomial. it's an equation! did you mean $x^3-3x-1$?

YES¡¡ $x^3-3x-1$
• Jun 8th 2009, 09:15 PM
NonCommAlg
Quote:

Originally Posted by osodud

Show that a polynomial of order 2 or 3 is irreducible on a field $F$ if has a root in $F$.

Quote:

Use this to show that the polynomial $x^3-3x=1$ is irreducible in $Z[x]$

Eisenstein's creiterio ??
since your polynomial is monic and of degree 3, in order to show that it's irreducible over $\mathbb{Z}$ we only need to prove that it has no root in $\mathbb{Z},$ i.e. the equation $x^3-3x-1=0$ has no integer

solution. suppose it has integer solutions. then $x(x^2-3)=1$ and hence we must have both $x=\pm 1$ and $x^2-3=\pm 1,$ which cannot happen and you're done.
• Jun 8th 2009, 09:21 PM
osodud
Quote:

Originally Posted by NonCommAlg

If and only if
• Jun 8th 2009, 09:25 PM
NonCommAlg
Quote:

Originally Posted by osodud
Show that a polynomial of order 2 or 3 is irreducible on a field $F$ if has a root in $F$.

Use this to show that the polynomial $x^3-3x=1$ is irreducible in $Z[x]$

Eisenstein's creiterio ??

that irreducible and has should be changed to either "reducible" and "has" or "irreducible" and "does not have".
• Jun 8th 2009, 09:53 PM
osodud
Final
Show that a polynomial of order 2 or 3 is reducible in a Field $F$ if and only if has one root in $F$.

That is what the exercise says.
I swear¡¡

If the exercise is incorrect could you tell me why is so??

you rule

Thanks again
• Jun 8th 2009, 10:22 PM
NonCommAlg
Quote:

Originally Posted by osodud

Show that a polynomial of order 2 or 3 is reducible in a Field $F$ if and only if has one root in $F$.

That is what the exercise says.
I swear¡¡

If the exercise is incorrect could you tell me why is so??

you rule

Thanks again

it is correct now. one side of the statement holds for any polynomial of degree at least 2: if a polynomial of degree at least 2 has a root $c \in F,$ then it has to be divisible by $x-c$ and thus it

is reducible. conversely, suppose $p(x) \in F[x]$ is a reducible polynomial of degree 2 or 3. then it can be factored into non-constant polynomials of smaller degrees. but how can you factor a

polynomial of degree 2 into non-constant polynomials of smaller degrees? there's one way only: two polynomials of degree 1. so suppose $p(x)=(ax+b)(cx+d).$ then obviously $x=\frac{-b}{a} \in F$

(and $x=\frac{-d}{c} \in F$) would be a root of $p(x).$ now suppose that the degree of $p(x)$ is 3. how can $p(x)$ be factored into non-constant polynomials of smaller degrees? there are two ways: three

polynomials of degree 1 or one polynomial of degree 1 and one polynomial of degree 2. in either case $p(x)$ has a factor of degree 1, say $ax+b,$ and thus $x=\frac{-b}{a} \in F$ would be a root of $p(x).$