Show that $f(x)=x^2+x+4$ is irreducible over $\mathbb{Z}_{11}$

Question

I know this can be done by evaluating $f$ at the points $0,1,...10$ to check if $f$ has a linear factor. Is there any other shorter way?

Use the quadratic formula, and see you need the square root of a quadratic nonresidue modulo $11$. — Angina Seng, Jun 06 '17 at 17:26
Or compute the $\gcd(x^{11}-x, x^2+x+4)$. If it had a root in $\mathbb{F}_{11}$ it would have a common root with $x^{11}-x$. (Probably not faster for this particular case since $11$ is so small) — Tob Ernack, Jun 06 '17 at 17:28
I get that roots must be (10x+√7)/2, (10x-√7)/2.But there exists no a, 0<=a<=10, such that a^2=7.Am i right? — jimm, Jun 06 '17 at 17:31
@AnwitaBhowmik You are working on $Z_{11}$! There is not square root of 7. — Our, Jun 06 '17 at 17:31
Or write $x^2+x+4\equiv(x+6)^2+1\pmod{11}$ and show that $-1$ is not a square in $\Bbb{Z}/11\Bbb{Z}$. — Servaes, Jun 06 '17 at 17:33
You can simply show that this polynomial is irreducible on $Z_3$, so it will directly follow that it is also irreducible on $Z_{11}$ — Our, Jun 06 '17 at 17:34
@Leth this plainly does not follow. Consider $g=x^2+7$ for example. It is irreducible in $(\Bbb{Z}/3\Bbb{Z})[x]$ but not in $(\Bbb{Z}/11\Bbb{Z})[x]$. — Servaes, Jun 06 '17 at 17:34
@Leth i meant that the roots must be (10x+a)/2, (10x-a)/2,where 0<=a<=10, such that a^2=7. — jimm, Jun 06 '17 at 17:37
@Servaes You are right, I have checked also the theorem, and to apply that, the coefficient must be integer. — Our, Jun 06 '17 at 17:38
The comment by @TobErnack uses the fact that $\prod_{a\in\Bbb{Z}/11\Bbb{Z}}(x-a)=x^{11}-x$, which requires some proof/argument/theory. And the computation of the gcd is also some work. I would say it is not the way to go if you're only just getting to know this material. — Servaes, Jun 06 '17 at 17:38
Yes in fact I am not quite sure the algorithm would really run faster than trying all values... — Tob Ernack, Jun 06 '17 at 17:39
@Servaes does this hold since 11 is a prime?I mean,can this result be used for Z/nZ for any n? — jimm, Jun 06 '17 at 17:45

score 2 · Accepted Answer · answered Jun 06 '17 at 18:20

2

We start with $$x^2+x+4=0$$ multiplying with $4$ we get $$4x^2+4x+16=0$$ This can be written as $$(2x+1)^2+15=0$$ modulo $11$ this is equivalent to $$(2x+1)^2-7=0$$

So we have to solve $$(2x+1)^2\equiv 7\mod 11$$

Using the quadratic reciprocity law we can calculate the legendre symbol

$$(\frac{7}{11})=-(\frac{11}{7})=-(\frac{4}{7})=-1$$

Hence $u^2\equiv 7\mod 11$ is not solveable because $7$ is not a quadratic residue modulo $11$. Since $x^2+x+4$ has no root in $\mathbb Z_{11}$, it follows that $x^2+x+4$ is irreducible in $\mathbb Z_{11}$

answered Jun 06 '17 at 18:20

Peter

84,454

I admit, that this method is not much shorter than plugging in the $11$ possible values, but the method will be efficient if the modulus is large. – Peter Jun 06 '17 at 18:22
3

The magically emerging $7$ is of course the discriminant $1^2-4\cdot 4$, so you might have arrived at that "at once". – Hagen von Eitzen Jun 06 '17 at 18:40
Can the quadratic formula be used for quadratic polynomials over any Zp(p being prime)? – jimm Jun 07 '17 at 04:17
Not for $p=2$ of course. – MooS Jun 07 '17 at 07:39

score 0 · Answer 2 · answered Jun 06 '17 at 19:11

0

In ${\mathbb Z}_{11}$, the discriminant of $x^2 + x + 4$ is $1 - 4 \cdot 4 = 7$, so this polynomial has a root in ${\mathbb Z}_{11}$ if and only if $7$ is a square modulo $11$. Using the law of quadratic reciprocity, $$\left(\frac{7}{11}\right) = -\left(\frac{11}{7}\right) = -\left(\frac{4}{7}\right) = -1,$$ so $7$ is not a square modulo $11$.

(This is essentially Peter's method, just jumping directly to the fact that you have to check whether or not $7$ is a quadratic residue modulo $11$.)

answered Jun 06 '17 at 19:11

Magdiragdag

15,049

Since you doubled an existing answer and only made it slightly simpler, let me do the same to you (while withholding posting an other answer): Using quadratic reciprocity here is an overkill, because $7=-4=-1 \cdot 2^2 \mod{11}$, i.e. $7$ is a square iff $-1$ is and this is not the case, because $11 \equiv 3 \mod 4$. – MooS Jun 06 '17 at 19:17

Will Jagy · Answer 3 · 2017-06-06T18:50:54.030

The discriminant of the quadratic form $u^2 + uv + 4 v^2$ is $-15.$ $$ (-15|11) = (-4|11) = (-1|11) (4|11) = (-1|11)=-1, $$ the last part because $11 \equiv 3 \pmod 4.$

Therefore, if $$ u^2 + uv + 4 v^2 \equiv 0 \pmod {11}, $$ it follows that both $u,v$ are divisible by $11.$ General theorem with proof at Prime divisors of $k^2+(k+1)^2$

So, with integer $x,$ $x^2 + x + 4$ is the same as $x^2 + xy + 4 y^2$ restricted to $y=1;$ that is, this $y$ is not divisible by $11.$ So $x^2 + x + 4$ cannot be divisible by $11$

I like this way of writing things, in quadratic forms this behavior is called "anisotropic." My experience is that existence of a solution that, however, violates some condition, is easier for the student to deal with than "infinite descent" arguments. All is the same, just a matter of emphasis.

Sigh. Same thing: $x^2 + 1$ is not divisible by any prime $q \equiv 3 \pmod 4.$ The discriminant of $u^2 + v^2$ is $-4.$ For such $q,$ if $q | (u^2 + v^2)$ then both $q|u$ and $q | v.$ However, it is not possible to have $q | 1,$ so $q$ cannot divide $x^2 + 1$ either.

Show that $f(x)=x^2+x+4$ is irreducible over $\mathbb{Z}_{11}$

3 Answers3