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Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.

The problem can be solved by putting every element from $\mathbb{Z}_{11}$ and checking if there is any root.

$0^2 + 0 + 4 = 4 \\ 1^2 + 1 + 4 = 6 \\ 2^2 + 2 + 4 = 10 \\ 3^2 + 3 + 4 = 5 \\ 4^2 + 4 + 4 = 2 \\ 5^2 + 5 + 4 = 1 \\ 6^2 + 6 + 4 = 2 \\ 7^2 + 7 + 4 = 5 \\ 8^2 + 8 + 4 = 10 \\ 9^2 + 9 + 4 = 6 \\ 10^2 + 10 + 4 = 4 $

But this process is very long one. I am seeking a short method by which I can solve such problem.

Now suppose that it is reducible, then the factors should be linear. So how can we solve the problem based on this kind of observation? Any help will be great. Thanks.

user26857
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Kushal Bhuyan
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    Compute the discriminant, see if it is a square $\pmod {11}$. – lulu Oct 31 '16 at 13:39
  • It is possible that this system would not work for other degrees than $2$. For instance a polynomial of degree $4$ could be the product of two irreducible quadratic polynomials and yet have no roots in the base ring. – Marc Bogaerts Oct 31 '16 at 17:44
  • $$\left(\frac{1-16}{11}\right) = \left(\frac{-1}{11}\right) = -1.$$ Done. – Jack D'Aurizio Oct 31 '16 at 21:42

3 Answers3

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By completing the square we get $$x^2+x+4=(x+6)^2+1.$$ So the original question reduces to the question if $-1$ is a square in $\mathbb Z_{11}$. This question can be answered with no using the first supplement for the law of quadratic reciprocity since $11 \equiv 3 \pmod 4$.

principal-ideal-domain
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As we can use the formula for the roots of a quadratic over any field with characteristic $\;\neq2\;$, calculating this quadratic's discriminant we get

$$\Delta=1-4\cdot4=-15=-4=7\pmod{11}$$

and now check the squares modulo $\;11\;$ :

$$0^2=0\;,\;\;1^2=1\;,\;\;2^2=4\;,\;\;3^2=9\;,\;\;4^2=5\;,\;\;5^2=3\;,\;\;6^2=3$$

and that is enough (why?), and thus there exists no $\;x\in\Bbb F_{11}\;$ with $\;x^2=7\;$ so the quadratic has no roots in $\;\Bbb F_{11}\;$ and is thus irreducible.

We could, of course, also use quadratic reciprocity instead of calculating the squares modulo $\;11\;$ (this is highly advisable for big primes) :

$$\left(\frac7{11}\right)\stackrel{Q.R}=-\left(\frac47\right)=-1$$

DonAntonio
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${\rm mod}\ 11\!:\,\ 0\equiv x^2\!+\!x\!+\!4\equiv x^2\!+\!12x\!+\!4\equiv (x\!+\!6)^2\!+1\,\Rightarrow\,(x\!+\!6)^2\equiv -1.$

But $\,-1\,$ is not a square, else by little Fermat: $\left[ a^2\equiv -1\right]^5\Rightarrow\, 1\equiv -1.$

Remark $ $ We implicitly applied Euler's Criterion to show $-1$ is not a square $\!\bmod 11.\,$ More generally we can use Legendre or Jacobi symbols or the generalized Euler Criterion.

Bill Dubuque
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