2

There is a similar question like this here

but I don't understand the solution. Since this is a fourth degree and since it has no root in $F_5$, it can only have a quadratic reduction. But how do we rule out the possibility that $x^4 + 2 = (x^2 + ax + b)(x^2 + cx + d)?$ Where the quadratic is irreducible in $F_5$?

There are a lot of similar questions like this on my practice exams.

Thanks

Lemon
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    $-2$ has order $4$ so a root $a$ has order $16$ thus $F_5(a) =F_{5^n}$ where $n$ is the least integer such that $16 | 5^n-1$. That is $[F_5(a):F_5]=n=4$, the minimal polynomial of $a$ has degree $4$ and hence it is $x^4+2$. – reuns Jun 29 '20 at 05:25
  • You can also try brute force method if you have time: there are a finite number of polynomials of degree $2$. Even fewer monic irreducible. – take008 Jun 29 '20 at 05:29
  • @take008 no this is an exam question, brute force takes too long. There are 5 choices for each $a,b,c,d$, that's too many – Lemon Jun 29 '20 at 05:32
  • @reuns sorry I haven't brushed up in a while, but how do we know $F_5(a) = F_{5^n}$? The left hand side is the simple extension right? – Lemon Jun 29 '20 at 05:34
  • Yes, $F_5(a)$ is a finite field containing $F_5$, so it is a $F_5$ vector space of finite dimension $n$ : it has $5^n$ elements. The next step is to prove that this field is the splitting field of $x^{5^n}-x$ thus unique (up to isomorphism) so we can call it $F_{5^n}$. – reuns Jun 29 '20 at 05:38

3 Answers3

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If we defined the Frobenius map by $F(a)=a^5$, then $a$ lies in $\Bbb F_{5^k}$ iff $F^k(a)=a$.

Here let $a$ be a root of $x^4+2=0$. Then $a^4=-2$ and $F(a)=a^5=-2a\ne a$. So $a\notin\Bbb F_5$.

Then $F^2(a)=F(-2a)=(-2a)^5=4a=-a\ne a$. Therefore $a\notin\Bbb F_{25}$.

Then $F^3(a)=F(-a)=(-a)^5=2a\ne a$. Therefore $a\notin\Bbb F_{125}$.

Of course, $F^4(a)=F(2a)=-4a=a$, so $a\in\Bbb F_{625}$.

Therefore $a$ generates the field extension of degree $4$ over $\mathbb{F}_5$, and so $X^4+2$ is ireeducible.

JackpotWizard 180
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Angina Seng
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  • why do we need to consider $F^2(a)$? Where did this come from? – Lemon Jun 30 '20 at 04:28
  • @Lemon Well, he is trying to find the $k$ such that $F^k(a)=a$, by the first line. So when $k=1$ didn't work, he tried $k=2$ and so on (know the comment is old, just to clearify for further users) – JackpotWizard 180 Mar 09 '24 at 20:15
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Assume $x^4+2$ factors in $\mathbb{Z}_5[x]$ as $$x^4+2=(x^2+ax+b)(x^2+cx+d)$$ If the $\text{RHS}$ is expanded,

  • The coefficient of the $x^3$ term is $a+c$, hence we must have $c=-a$.$\\[4pt]$
  • The constant term is $bd$, hence we must have $d={\large{\frac{2}{b}}}$.

hence the factorization can be rewritten as $$x^4+2=(x^2+ax+b)(x^2-ax+\frac{2}{b})$$ Expanding, we get $$x^4+2=x^4+\left(b+\frac{2}{b}-a^2\right)x^2+\left(\frac{2a}{b}-ab\right)x+2$$ hence we must have $$ \left\lbrace \begin{align*} &\frac{2a}{b}-ab=0&&(\text{eq}1)\\[4pt] &b+\frac{2}{b}-a^2=0&&(\text{eq}2)\\[4pt] \end{align*} \right. $$ If $a\ne 0$, $(\text{eq}1)$ yields $b^2=2$, and if $a=0$, $(\text{eq}2)$ yields $b^2=-2$.

Either way we have a contradiction since $2$ and $-2$ are not squares in $\mathbb{Z}_5$.

quasi
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Here's another proof . . .

Assume $x^4+2$ factors in $\mathbb{Z}_5[x]$ as $$x^4+2=(x^2+ax+b)(x^2+cx+d)$$ Let $K$ be an algebraic closure of $\mathbb{Z}_5$, and let $r$ be a root in $K$ of the polynomial $x^4+2$.

Then identically we have $$r^4=(-r)^4=(2r)^4=(-2r)^4$$ hence $\pm r,\pm 2r$ are roots in $K$ of $x^4+2$, and are distinct since $r\ne 0$.

Since $b$ is the product of the roots in $K$ of $x^2+ax+b$, it follows that $b$ is the product of two distinct elements of the set $\{\pm r,\pm 2r\}$, hence $b\in\{-r^2,\pm 2r^2,-4r^2\}$.

In any case, since $b\in\mathbb{Z}_5$, it follows that $r^2\in\mathbb{Z}_5$.

But $r^2\in\mathbb{Z}_5$ implies $r^4$ is a square in $\mathbb{Z}_5$, contradiction, since $r^4=-2$ which is not a square in $\mathbb{Z}_5$.

quasi
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  • where did 2r come from? – Lemon Jun 30 '20 at 04:30
  • @Hawk: Identically we have $$(2r)^4=16r^4=r^4$$ since $16=1$ in $\mathbb{Z}_5$. – quasi Jun 30 '20 at 06:42
  • There is quite a few lines of reasoning I am not following. Like why consider the equation $r^4 = (2r)^4$? Why not $r^4 = (-r)^4 = (4r)^4 = 256r^4 = 1r^4$? Why $2$? – Lemon Jun 30 '20 at 06:48
  • @Hawk: In any ring we have $(-r)^4=r^4$. But in the ring $Z^5$, we have $2^4=16=1$, hence, since $K$ contains $Z_5$ as a subfield, we have $(2r)^4=2^4r^4=16r^4=1r^4=r^4$. – quasi Jun 30 '20 at 07:06
  • @Hawk: Is there anything else I can help clarify with regard to my answer? – quasi Jul 03 '20 at 23:17
  • I don't feel like my other question was answered. It seems $4$ is more natural to choose than $2.$ I know why they are equal, but I want to know why you chose $2$. – Lemon Jul 03 '20 at 23:37
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    @Hawk: By Fermat's little Theorem, $j^4=1$ for all nonzero $j\in\mathbb{Z}^5$, hence identically we have $$r^4=(2r)^4=(3r)^4=(4r)^4$$ Then since $r$ is a root in $K$ of $x^4+2$, so are each of $2r,3r,4r$. To make it look nicer, I chose to write $3r$ as $-2r$ and $4r$ as $-r$. – quasi Jul 04 '20 at 00:07
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    @Hawk: Thus in $K[x]$, the polynomial $x^4+2$ factors as $$x^4+2=(x-r)(x-2r)(x-3r)(x-4r)$$ or equivalently $$x^4+2=(x-r)(x-2r)(x+2r)(x+r)$$ so $$(x^2+ax+b)(x^2+cx+d)=(x-r)(x-2r)(x+2r)(x+r)$$ hence the product of two of the four factors of the $\text{RHS}$ must identically be equal to $x^2+ax+b$ and the product of the other two factors must identically be equal to $x^2+cx+d$. It follows that the value of $b$ must be one of $-r^2,2r^2,-2r^2,-4r^2$. – quasi Jul 04 '20 at 03:00
  • @Hawk: Is the logic clearer now? – quasi Jul 04 '20 at 03:01
  • yes a bit better. how do we deduce $r^2\in Z_5 \implies r^4$ is a square in $Z_5$ – Lemon Jul 04 '20 at 06:31
  • @Hawk: $r^4=(r^2)^2$. – quasi Jul 04 '20 at 07:37
  • No but i mean IN $Z_5$. We only know that $a^5 = a$ for every $a \in Z_5$ from Fermat. – Lemon Jul 04 '20 at 07:54
  • Other than testing every number in $Z_5$. What properties in $Z_5$ preserves this squaring? Actually is it just group closure? – Lemon Jul 04 '20 at 07:56
  • @Hawk: In any ring $R$, if $w\in R$, then $w{,\cdot,}w$ is a square in $R$, hence once we know $r^2\in\mathbb{Z}_5$, it follows that $r^2{,\cdot,}r^2$ is a square in $\mathbb{Z}_5$, so $r^4$ is a square in $\mathbb{Z}_5$. – quasi Jul 04 '20 at 08:17
  • @Hawk: As regards the squares in $\mathbb{Z_5}$, since $\mathbb{Z_5}$ only has $5$ elements, there's no need to invoke anything more sophisticated than evaluating $$0^2=0,;;;;;1^2=1,;;;;;2^2=4,;;;;;3^2=(-2)^2=4,;;;;;4^2=(-1)^2=1$$ so the squares in $\mathbb{Z}_5$ are $0,1,4$. In particular, $2$ and $-2$ are not squares in $\mathbb{Z}_5$. – quasi Jul 04 '20 at 08:28