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Reduction into linear factors $\mathbb{Z}_{17}[x]$:

This part is not too hard: $x^4 \equiv -1$ mod 17 has solutions: 2, 8, 9, 15 so

$(x-2)(x-8)(x-9)(x-15) = x^4 -34 x^3 +391 x^2-1734 x+2160 \equiv x^4+1$ mod 17.

Reduction over $\mathbb{Z}_{11}[x]$:

This one doesn't have such an easy solution, as neither of $y^2 \equiv -1$ mod 11 or $x^4 \equiv -1$ mod 11 have solutions.

I've tried $x^4+1 = (x^2- \sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ but $x^2 \equiv 2$ mod 11 also has no solutions so there's no easy substitution here.

I think that this approach is not going to work here, so I need something new. Any suggestions?

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    $\begingroup$ See this question for a more or less complete answer. $\endgroup$ – Jyrki Lahtonen Apr 5 '13 at 15:05
  • $\begingroup$ It certainly has a very complete answer - but perhaps somewhat overkill when it comes to answering this question? Nonetheless thanks for the link. $\endgroup$ – Lewy Apr 5 '13 at 15:19
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    $\begingroup$ You were almost there... You tried $X^4+1=X^4+1+2X^2-2X^2$. Now, if you observe that $2=-9$ you are done.... $\endgroup$ – N. S. Apr 5 '13 at 15:45
  • $\begingroup$ The splitting modulo $p=17$ comes from the fact that $p \equiv 1 \pmod 8$, so that $p$ is totally split in the extension $\Bbb Q(\zeta_8) / \Bbb Q$, which is the splitting field of $x^4+1$. $\endgroup$ – Watson Apr 6 '18 at 13:30
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I will assume you've made no mistakes in your work.

Since $-1$ hasn't a fourth root in $\Bbb Z_{11},$ then the only way to reduce $x^4+1$ over $\Bbb Z_{11}$ is as a product if quadratics. We may as well assume that the quadratics are monic, so we've got $$x^4+1=(x^2+ax+b)(x^2+cx+d),$$ leaving us with the following system: $$a+c=0\\ac+b+d=0\\ad+bc=0\\bd=1$$ By the first equation, $c=-a,$ and the rest of the system becomes: $$b+d=a^2\\a(d-b)=0\\bd=1$$ Now, if $a=0$, then the first and third remaining equations imply that $-1$ has a square root in $\Bbb Z_{11}$ (why?), which you've ruled out. Hence, we need $b=d,$ and the rest of the system becomes: $$2b=a^2\\b^2=1$$ Can you take it from here?

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  • $\begingroup$ Yes, thank you for your clear explanation. This gives two possible solutions: $(x^2-3x -1)(x^2 +3x-1)$ and $(x^2-8x -1)(x^2 +8x-1)$. $\endgroup$ – Lewy Apr 5 '13 at 15:27
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    $\begingroup$ @Lewy: You do know that factorization in this ring is unique, don't ya? IOW, look more closely at the "two" factorizations :-) $\endgroup$ – Jyrki Lahtonen Apr 5 '13 at 15:31
  • $\begingroup$ @Lewy: Jyrki's point is well-taken. $8=-3,$ so those factorizations are identical (up to arrangement of the factors). $\endgroup$ – Cameron Buie Apr 5 '13 at 15:51
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    $\begingroup$ Yes, I see that. Perhaps this is one of those times at which it's appropriate to say "Doh!". $\endgroup$ – Lewy Apr 5 '13 at 15:55
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An abstract algebra argument, without much computation.

Claim: The polynomial $x^4+1$ splits over $\mathbb F_{p^2}$ for any odd (*) prime $p$.

We can see this because $F_{p^2}^\times$ is a cyclic group of order $p^2-1$, a multiple of $8$. If $g$ is a generator, then $g^{(1+2k)\frac{p^2-1}{8}}$ gives distinct roots for $k=0,1,2,3$.

This means that $x^4+1$ can't be irreducible in $\mathbb Z_p$, because if it was, the splitting field would have $p^4$ elements.

(*) True for $p=2$, but not for the same reasons. When $p=2$, $x^4+1=(x+1)^4$.

Specific solutions:

We can make this explicit by looking at the "standard" complex $4$th roots of $-1$: $$\pm\frac{\sqrt 2}2 \pm\frac{\sqrt{2}}{2}i=\frac{1}{2}(\pm \sqrt{2}\pm\sqrt{-2})$$

If $p\equiv 1\pmod 8$ then $2$ and $-2$ have square roots in $\mathbb Z_p$, so $x^4+1$ has four roots.

If $p\equiv -1\pmod 8$ then $2$ is a square, but $-2$ is not. Write $a^2\equiv 2\pmod p$. Then write:

$$x^4+1 = \left(x^2 -ax + 1\right)\left(x^2+ax+1\right)$$

If $p\equiv 3\pmod 8$ then $-2$ is a square and $2$ is not. Letting $b^2\equiv -2\pmod p$, we get:

$$x^4+1 = \left(x^2-bx-1\right)\left(x^2+bx-1\right)$$

Finally, if $p\equiv 5\pmod 8$ then neither of $\pm 2$ is a square, but $-1$ is a square. Letting $c^2\equiv -1\pmod p$, we see that the roots are $(\pm 1 \pm c)\sqrt{2}/2$ and the result is:

$$x^4+1 = (x^2-c)(x^2+c)$$

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    $\begingroup$ The use of the formula $(\pm\sqrt2\pm\sqrt{-2})/2$ is a nice trick. In the linked answer I (IIRC at least partly in response to the discussion in the comments that lead to CWification) did get there, but should have seen it right away. +1 for unveiling the reason for the emergence of $\sqrt{\pm2}$ to this extent. $\endgroup$ – Jyrki Lahtonen Apr 5 '13 at 17:42
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    $\begingroup$ Perhaps a more direct proof that $x^4+1$ is not irreducible is just to note that $x^4+1|x^8-1|x^{p^2-1}-1|x^{p^2}-x$. And $x^{p^2}-x$ has as its only irreducible factors the quadratic and linear prime polynomials. $\endgroup$ – Thomas Andrews Apr 5 '13 at 18:04
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    $\begingroup$ Well, that's exactly how I did it originally. But somewhat surprisingly people don't think this is a duplicate of that :-) $\endgroup$ – Jyrki Lahtonen Apr 5 '13 at 18:07
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    $\begingroup$ @JyrkiLahtonen N.S. suggested in comments to the question: $x^4+1=x^4+1 +2x^2-2x^2$. Hahaha, that makes the primacy of $\sqrt{2}$ and $\sqrt{-2}$ even more obvious! $\endgroup$ – Thomas Andrews Apr 5 '13 at 18:16
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This question in fact already answered your question. Indeed, in the accepted answer, the complete solution can be found. So let me explain this and put it into CW.
As expounded in the link, we know that there is a primitive root of order $8$ in $\mathbb F:=\mathbb F_{p^2}$, called $u$. Since $\mathbb F$ is of degree $2$ over $\mathbb F_p$, the minimal polynomial is of degree at most $2$ over $\mathbb F_p$. Here, our $p\equiv 3\pmod 8$, so there is no $u$ in $\mathbb F_p$, as $u^4=-1$. Its conjugate is then $u^p=u^3$. Therefore $x^4+1$ has a factor given by $(x-u)(x-u^3)$. After easy calculations, we find that this becomes $(x^2-(u+u^3)x-1)$. And the other factor of that one must, after easy calculations of other conjugates of $u$, be $x^2+(u+u^3)x-1$. Therefore, $a:=u+u^3$ satisfies $a^2+2=0$. And I guess you know there is one such $a$ for $p=11$.

For those who jumped here:
We found the factorisation: $$(x^2-3x-1)(x^2+3x-1)$$.

Thanks for the attention, and inform me of any errors. Thanks.

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    $\begingroup$ Also see that referred answer for more detail. $\endgroup$ – awllower Apr 5 '13 at 15:21
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Look for a factorization of the shape $(x^2-ax+b)(x^2+ax+b)$. So we want $b^2\equiv 1\pmod{11}$, with $2b$ a quadratic residue.

Remark: Note that this is a general procedure for $x^4+1$ modulo an odd prime. The issue is whether one of $2$ or $-2$ is a quadratic residue of $p$. One of them is unless $p\equiv 5\pmod{8}$.

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    $\begingroup$ In the case $p\equiv 5\pmod8$ we have that $2$ is a quadratic non-residue, so $i:=2^{(p-1)/4}$ is a square root of $-1$. In that case we have the factorization $(x^2+i)(x^2-i)$. $\endgroup$ – Jyrki Lahtonen Apr 5 '13 at 15:35
  • $\begingroup$ In general, at least one of $a,b,ab$ is a square $\pmod p$. Take $a=2, b=-1.$ :) $\endgroup$ – Thomas Andrews Apr 5 '13 at 16:12
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    $\begingroup$ Note, the standard formulation for the complex roots of $x^4+1=0$ can be formulated as $\frac{1}{2}(\pm\sqrt{2}\pm\sqrt{-2})$, which is why these two roots are important. $\endgroup$ – Thomas Andrews Apr 5 '13 at 16:14
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Hint: We know that if $f(x)=x^4+1$ is reducible, then it either has a root or can be written $f(x)=g(x)h(x)$ where $\deg(g)=\deg(h)=2$.

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  • $\begingroup$ I guess OP already knew this, judged from the try. The point is: how to find those degree $2$ factors? $\endgroup$ – awllower Apr 5 '13 at 15:03
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You may use the fact that $$x^2 = 9 \mod 11$$ has a solution, say $x = c$. Then use $$(x^2 + cx - 1)(x^2 - cx - 1),$$ and $c$ is pretty much obvious from here...

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