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Factor $x^4-2$ as a product of irreducible polynomials in $\mathbb{Z}_3[x]$, $\mathbb{Z}_{17}[x]$, $\mathbb{Q}[x]$, $\mathbb{R}[x]$, and $\mathbb{C}[x]$.

I think I figured out $\mathbb{R}[x]$ with $(x^2 + \sqrt{2})(x-\sqrt[4]{2})(x+\sqrt[4]{2})$, I don't think the first term can be factored any further.

I also got none for $\mathbb{Q}[x]$.

For $\mathbb{C}[x]$ I got $(x^2 - 2i)(x^2 - 2i)$.

$\mathbb{Z}_3[x]$ and $\mathbb{Z}_{17}[x]$, I got none from plugging in all their respective numbers and not getting 0 for any of them.

Could someone let me know if this is right? I am unsure if I am missing something for $\mathbb{Q}[x]$, $\mathbb{Z}_3[x]$ and $\mathbb{Z}_{17}[x]$ if it is reducible.

user973229
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    For the first one, did you mean: $(x^2 + \sqrt{2})(x-\sqrt[4]{2})(x+\sqrt[4]{2})$? – Shooting Stars Dec 02 '21 at 23:32
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    Remember for $\mathbb{Q}$, you can prove this using Eisenstein's criterion. – briemann Dec 02 '21 at 23:48
  • Check you answer for reducibility over $\mathbb{C}[x]$. Does $(x - 2i)(x - 2i) = x^4 - 2$ ? Also consider the irreducibility of $(x^2 + \sqrt{2})(x-\sqrt[4]{2})(x+\sqrt[4]{2})$ over $\mathbb{R}[x]$. Does it now become reducible over $\mathbb{C}[x]$?... Just because its over the complex domain now, this does not mean you cannot have full real factors. I.E. you do not need all your factors over $\mathbb{C}[x]$ to include imaginary numbers. Remember that $\mathbb{R} \in \mathbb{C}$ – Dstarred Dec 02 '21 at 23:59
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    @ShootingStars ah yes, I've fixed it now – user973229 Dec 03 '21 at 07:31
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    @UnexpectedConfusion Oh, I see, so for $\mathbb{C}[x]$, does that then become $(x+i \sqrt[4]{2})(x - i \sqrt[4] {2}) (x-\sqrt[4]{2}) (x+\sqrt[4]{2})$? – user973229 Dec 03 '21 at 07:39
  • Indeed thats correct – Dstarred Dec 03 '21 at 08:13

2 Answers2

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You had a wrong sign in the factorization in $\mathbb{R}[x]$ (typo): $$ (x^2+\sqrt{2}\,)(x-\sqrt[4]{2}\,)(x+\sqrt[4]{2}\,) $$ is the factorization: the quadratic factor has no roots in $\mathbb{R}$, so it's irreducible.

The polynomial is indeed irreducible in $\mathbb{Q}[x]$: it has no rational root so it would be a product of two quadratic factors, but such a decomposition would also lead to a splitting in $\mathbb{R}[x]$ and the only way you can combine the previous factorization in the required way would be $(x^2+\sqrt{2}\,)(x^2-\sqrt{2}\,)$ and the factors are not in $\mathbb{Q}[x]$. Or you can use Eisenstein's criterion, which applies to $x^4-2$ for the prime $2$.

Sorry, but the factorization in $\mathbb{C}[x]$ is wrong: it should be a product of linear factors and you get them from the factorization in $\mathbb{R}[x]$.

I'll do the case $\mathbb{Z}_3[x]$. The polynomial has no roots in $\mathbb{Z}_3$, but it could be a product of two (monic) quadratic factors: $$ x^4-2=(x^2+ax+b)(x^2+cx+d) $$ Since $bd=-2=1$, you can have $b=1$ and $d=1$ or $b=2$ and $d=2$. In the first case we have $$ (x^2+ax+1)(x^2+cx+1)=x^4+(a+c)x^3+(2+ac)x^2+(a+c)x+1 $$ and so you need $a+c=0$ and $ac+2=0$. Hence $c=-a$ and $a^2=2$, which is impossible.

In the second case we have $$ (x^2+ax+2)(x^2+cx+2)=x^4+(a+c)x^3+(1+ac)x^2+(2a+2c)x+1 $$ Again $a+c=0$ and $1+ac=0$. Since $c=-a$, we obtain $a^2=1$. Well, this is indeed possible: we have $a=1$ or $a=2$. Thus the factorization is $$ (x^2+x+2)(x^2+2x+2) $$

egreg
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  • One way to know in advance that the polynomial will factorize over $\mathbb{Z}_3$ is to note that in $GF(9)$ every non-zero element has order $8$, so in particular we already have all $4$ roots of $x^4=-1=2$. However if $x^4-2$ was irreducible over $\mathbb{Z}_3$, its splitting field would be degree $4$ over $\mathbb{Z}_3$, not degree $2$. Of course this does not give you the factorisation - just tells you it exists, so (+1) for getting it. – tkf Dec 03 '21 at 00:35
  • @egreg my bad, yes it was a typo. Okay, for $\mathbb{C}[x]$, I was able to get $(x+i \sqrt[4]{2})(x - i \sqrt[4] {2}) (x-\sqrt[4]{2}) (x+\sqrt[4]{2})$. Thank you, I kind of understand your explanation for $\mathbb{Z}_3[x]$. – user973229 Dec 03 '21 at 08:18
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The $\mathbb{Z}_{17}$ case remains. The units of $\mathbb{Z}_{17}$ form a cyclic group of order $16$. The order of $2$ is $8$, as $2^4=-1$. Thus $\sqrt{2}\in \mathbb{Z}_{17}$, and your polynomial factorises as: $$x^4-2=(x^2+\sqrt{2})(x^2-\sqrt{2}).$$

It cannot factorise further as if there were a linear factor, then $\mathbb{Z}_{17}$ would contain an element $\alpha$ with $\alpha^4=2$, so $\alpha^{16}=-1$, which is impossible as for any element $\alpha \in \mathbb{Z}_{17}$ we have $\alpha^{16}=1$.

To complete the question you should work out $\sqrt{2}$ in $\mathbb{Z}_{17}$. The easiest way to do this is check values of $2+17k$, for $k=0,1,2,3,\ldots$ till you get a perfect square.

tkf
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  • Possibly too advanced for the OP, but it's good to see different attack methods. – egreg Dec 03 '21 at 07:51
  • Could you explain a bit more about the statement that "the order of 2 is 8?" I understand that $2^4 = -1$ in $\mathbb{Z}{17}$ though I don't understand how $\sqrt{2} \in \mathbb{Z}{17}$ where I thought all the elements in $\mathbb{Z}_{17}[x]$ were the integer $0,1,...,16$. For $2+17k$, I got a perfect square with $k = 2$ and $k = 7$ though I don't know how to proceed from here. – user973229 Dec 03 '21 at 07:59
  • You got $k=2$. So $2+2*17=36=6^2$. Then $6^2=2 \mod 17$, so the square root of $2$ modulo $17$ is $6$. Substitute into the factorisation and you get $x^4-2=(x^2-6)(x^2+6)$. As I explained this does not factorise any further. – tkf Dec 03 '21 at 08:02
  • As for the order of $2$: We know $2^8=(-1)^2=1 \mod 17$, so the order of $2$ must divide $8$. However it cannot divide $4$, as $2^4\neq 1 \mod 17$. The only natural number which divides $8$ but does not divide $4$, is $8$. – tkf Dec 03 '21 at 08:05
  • @tkf this does go over my head a little but thank you, I appreciate your answer. – user973229 Dec 03 '21 at 20:35
  • Yes I suppose there are two bits of theory I use here: (1) Field theory: The multiplicative group of units of a finite field is always cyclic (which follows from a polynomial never having more roots than its degree). (2) Group theory: The order of an element always divides the order of the group etc. $${}$$ Still you should be able to check that $x^4−2=(x^2−6)(x^2+6)$ modulo $17$, and you already showed it does not factorise further by plugging in all the values. The only issue is knowing to look for a factorisation of that form, which saves a lot of time in $\mathbb{Z}_{17}$. – tkf Dec 03 '21 at 23:16