# Math Help - Irreducible Polynomial 2

1. ## 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 ??

2. 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 $x^3-3x-1$?

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

4. 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 ??
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.

5. Originally Posted by NonCommAlg
If and only if

6. 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".

7. ## 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

8. 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 $p(x).$