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Over which of the following fields $x^{4} - 2x^{3} + 2x^{2} - 2x + 1$ can be factorized to linear factors?

  1. $\mathbb{C}$ 2. $\mathbb{Q}$ 3. $\mathbb{Z_2}$ 4. $\mathbb{Z_3}$

There is no answer in the textbook, so I wanted to check if my solution is correct.

After linear factorization over reals I get the following factors: $(x-1)(x^{2}+1)(x-1)$

  1. Over $\mathbb{C}$, Yes, (x-1)(x-1)(x+i)(x-i)
  2. Over $\mathbb{Q}$, No, because $(x^{2}+1)$ doesn't have roots over $\mathbb{Q}$
  3. Over $\mathbb{Z_2}$, (I am not sure if that is a correct step) first, I convert $x^{4} - 2x^{3} + 2x^{2} - 2x + 1$ to $\mathbb{Z_2}$, so we get $x^{4}+1$, now there are factors $(x^{2}+1)(x^{2}-1)$, so the root is 1, as a check $1^{4}+1$ = 0, so Yes
  4. Over $\mathbb{Z_3}$, First, convert $x^{4} - 2x^{3} + 2x^{2} - 2x + 1$ to $\mathbb{Z_3}$, so we get $x^{4} + x^{3} + 2x^{2} + x + 1$, now I couldn't find any linear factorization here and checks for 0,1,2 did not reveal any roots, so No

Questions:

a) Is the step with converting polynomial to $\mathbb{Z_2}$ and $\mathbb{Z_3}$ correct or I completely misunderstood that? If not, what would be the correct way?

b) Is there any useful trick to quickly determine if polynomial of degree n irreducible/not linearly reducible in $\mathbb{Z_k}$?

Jill White
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  • $x=1$ should be a root in every field... Also, the polynomial is equivalent to $(x-1)^4+2(x-1)^2$, so that might help identify roots more easily. – abiessu Sep 02 '18 at 00:13
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    Why convert when you have the factorization as $(x-1)^2(x^2+1)$. This means you need to only check if -1 is a square in the field. in $\mathbb{C}$ and $\mathbb{Z}_2$ it is but in $\mathbb{Q}$ and $\mathbb{Z}_3$ it is not. – Marco Sep 02 '18 at 00:15
  • @Marco indeed...didn't think about it for some reason, thanks! – Jill White Sep 02 '18 at 00:28
  • take a look at this one. For inspiration. https://math.stackexchange.com/questions/2887405/set-of-primes-p-which-x4-x3-2x2-2x-1-completely-factors-in-finite-field-o – Will Jagy Sep 02 '18 at 00:33
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    Hmmm. That is actually a fair question for you to begin investigating, Jill. You have already considered primes $2$ and $3.$ Bigger question: for which primes $p$ does your polynomial factor completely in $\mathbb Z_p ; ?$ – Will Jagy Sep 02 '18 at 00:38
  • Note that @Marco has given a complete solution for all fields: Your polynomial factors into linear factors over a field if and only if that field contains a square root of $-1$. – Andreas Blass Sep 02 '18 at 01:22

1 Answers1

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Your solution is correct.

Note that you have $$ (x-1)^2(x^2+1)$$ so your only remaining concern about linear factorization is $x^2+1$ which factors on $\mathbb {C}$ and $\mathbb {Z_2}$ and does not factor over $\mathbb {Q}$ and $\mathbb {Z_3}$ because in $\mathbb {Z_2}$ we have $x^2+1=x^2-1$ which factors and in $\mathbb {Z_3}$ we have $(x^2+1)=x^2-2$ which does not factor.

Mohammad Riazi-Kermani
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