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

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

Eisenstein's creiterio ??

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- Jun 8th 2009, 07:59 AMosodudIrreducible Polynomial 2
Show that a polynomial of order 2 or 3 is irreducible on a field $\displaystyle F $ if has a root in $\displaystyle F$.

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

Eisenstein's creiterio ?? - Jun 8th 2009, 08:54 PMNonCommAlg
- Jun 8th 2009, 08:57 PMosodud
- Jun 8th 2009, 09:15 PMNonCommAlg
this is false. fix the mistake(s) you made and then i'll help you.

Quote:

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

Eisenstein's creiterio ??

__integer__

__solution__. suppose it has integer solutions. then $\displaystyle x(x^2-3)=1$ and hence we must have both $\displaystyle x=\pm 1$ and $\displaystyle x^2-3=\pm 1,$ which cannot happen and you're done. - Jun 8th 2009, 09:21 PMosodud
- Jun 8th 2009, 09:25 PMNonCommAlg
- Jun 8th 2009, 09:53 PMosodudFinal
Show that a polynomial of order 2 or 3 is reducible in a Field $\displaystyle F$ if and only if has one root in $\displaystyle F$.

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 PMNonCommAlg
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 $\displaystyle c \in F,$ then it has to be divisible by $\displaystyle x-c$ and thus it

is reducible. conversely, suppose $\displaystyle p(x) \in F[x]$ 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 $\displaystyle p(x)=(ax+b)(cx+d).$ then obviously $\displaystyle x=\frac{-b}{a} \in F$

(and $\displaystyle x=\frac{-d}{c} \in F$) would be a root of $\displaystyle p(x).$ now suppose that the degree of $\displaystyle p(x)$ is 3. how can $\displaystyle p(x)$ 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 $\displaystyle p(x)$ has a factor of degree 1, say $\displaystyle ax+b,$ and thus $\displaystyle x=\frac{-b}{a} \in F$ would be a root of $\displaystyle p(x).$