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Consider the polynomial $$f(x)=x^4-x^3+14x^2+5x+16$$ and $\mathbb{F}_p$ be the field with $p$ elements, where $p$ is prime. Then

  1. Considering $f$ as a polynomial with coefficients in $\mathbb{F_3}$, it has no roots in $\mathbb{F_3}$.
  2. Considering $f$ as a polynomial with coefficients in $\mathbb{F_3}$, it is a product of two irreducible factors of degree $2$ over $\mathbb{F_3}$.

  3. Considering $f$ as a polynomial with coefficients in $\mathbb{F_7}$, it has an irreducible factor of degree $3$.

  4. $f$ is a product of two polynomials of degree two over $\mathbb{Z}$.

My work: $$f(x)=x^4+2x^3+2x^2+2x+1$$ in $\mathbb{F_3}$ which can be written as $$f(x)=(x^2+1)(x+1)^2$$ hence $1$ and $2$ are wrong.

In $\mathbb{F_7}$, we can write

$$f(x)=x^4+6x^3+5x+2,$$ which is reducible but I am unable to conclude $3$ precisely and also having problem with $4$.

Am i right with my conclusions? Help me out.

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    $\begingroup$ $f(x) = x^4 - x^3 - 2x + 2 ~(\mod 7)$ Note that $f(1) = 0$ hence $x-1$ is ?. If we can prove point 3 then point 4 will be false since if $f$ is a product of two polynomials of degree 2 in $\mathbb{Z}$ then it should not has an irreducible factor of degree 3 in $\mathbb{Z}_7$ $\endgroup$ – AmerYR Oct 26 '15 at 4:05
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    $\begingroup$ @Ameryr: I encourage you to flesh that argument out to an answer (for parts 3 and 4). The way you wrote the polynomial modulo $7$ shows that $f(x)=(x-1)(x^3-2)$. Checking that that cubic factor is irreducible over $\Bbb{F}_7$ is easy. $\endgroup$ – Jyrki Lahtonen Oct 26 '15 at 6:50
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You can write $f(x) = x^4 - x^3 -2x + 2 ~(\mod 7)$ and you can check that $f(1) = 0$. Hence $(x-1)$ is a factor of $f(x) ~(\mod 7)$.

After factoring we get $f(x) = (x-1)(x^3-2) ~(\mod 7)$. So If we proved that $x^3 - 2$ is irreducible $\mod 7$ that means point $3$ is true and point $4$ is false since if $f(x)$ can be written as a product of polynomials of degree two then that hold for $f(x) ~(\mod 7)$ but this contradict that $f(x) $ has an irreducible factor of degree $3$.

To prove $x^3 - 2$ is irreducible we can check $1^3, 2^3, \cdots , 6^3 ~(\mod 7)$ if non equal to $2$ then we are done. And that can be proved easily using the theorem $a^{p-1} \equiv 1 ~(\mod p)$ for $(a,p)=1$. Thus $a^{\frac{p-1}{2}} \equiv \pm 1~ (\mod p)$.

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To see if the 4th degree polynomial over $\mathbb{Z}_7$ has an irreducible 3rd degree factor, simply check to see if it has a root in $\mathbb{Z}_7$ first by trying all $7$ options. If so, divide by the corresponding factor to leave you with a cubic. Now either the cubic is irreducible, or it has a root in $\mathbb{Z}_7$. Again, try all $7$ options to see.

Over $\mathbb{Z}$, you can look for linear factors by looking for integer roots. And you can look for those by applying the rational root theorem. If you find none, you can move on to trying to represent it as $(x^2+ax+b)(x^2+cx+d)$. You can use your factorization over $\mathbb{Z}_3$ to refine this as $(x^2+3ax+(3b+1))(x^2+(3c+2)x+(3d+1))$, if that helps. Multiply it out and you can slowly rule out all cases, or possibly find the asserted factorization.

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