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I am revising for a discrete mathematics exam and as quite stuck on this question.

Show that the polynomial $f = x^2 + 2 x + 3 \in \mathbb{Z}_5[x]$ is primitive. How many monic primitive quadratic polynomials are there in $\mathbb{Z}_5[x]$?

Any help would be greatly appreciated.

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Use LaTeX to properly write mathematics in this. Directions to be found in the FAQ section – DonAntonio Apr 11 '13 at 11:19
What do you call "primitive polynomial over a finite field" to, please? Could it be you actually meant "irreducible"? – DonAntonio Apr 11 '13 at 11:21
@DonAntonio: Primitive polynomial has a standard meaning in the context of finite fields. A polynomial is said to be primitive, if its zero generates the multiplicative group of the extension field. Makes them useful in building discrete logarithm tables and such applications. – Jyrki Lahtonen Apr 11 '13 at 12:20
Jyrki, I know that, but what extension are we talking about here? If the OP really wanted that then any irreducible polynomial over some field $\,\Bbb F\,$ with a root $\,w\,$ in some extension is primitive as it is the minimal polynomial of the primitive element $\,w\,$ in the simple extension$\,\Bbb F(w)/\Bbb F\,$ weird, imo, and I still am not sure whether the OP really meant something related to this – DonAntonio Apr 11 '13 at 12:28
No, @DonAntonio. Not all the irreducible polynomials are primitive in this sense. For example $x^2+x+1$ is irreducible over $\mathbb{F}_5$, but its roots are cubic roots of unity, and thus won't generate multiplicatively (sorry about not making that clear right away) the entire multiplicative group, in this case the cyclic group of order $5^2-1=24$. – Jyrki Lahtonen Apr 11 '13 at 12:40

A polynomial is called primitive (in the context of finite fields), iff its zero is a generator of the multiplicative group of the field it generates. In this case the polynomial is quadratic, so a root $\alpha$ will generate the field $L=\mathbb{F}_{25}$.

The multiplicative group of $L$ is cyclic of order 24. By the well known facts about cyclic groups, the group $L^{\times}$ has $\phi(24)=\phi(2^3\cdot3)=2^2(3-1)=8$ generators. Another way of seeing this is to observe that if $g$ is a primitive root of unity of order $24$, all of them are in the list $g$, $g^5$, $g^7$, $g^{11}$, $g^{13}$, $g^{17}$, $g^{19}$ and $g^{23}$. Anyway, between them, these eight elements have four distinct minimal polynomials (the conjugate of a primitive element is also primitive, and the conjugates come in pairs in a quadratic extension).

The answer is thus that there are exactly $4$ quadratic primitive polynomials.

Looking at the specific polynomial $p(x)=x^2+2x+3$. Its discriminant is non-square, so it is irreducible (alternatively one may check that it has no zeros in $\mathbb{F}_5$). Let $g$ be one of its roots in the extension field $\mathbb{F}_{25}$. We want to prove that it is of order $24$. We know that the order is a factor of $24$, so it suffices to rule out factors of $12$ and $8$ as orders.

The other zero of $p(x)$ is the Frobenius conjugate $g^5$ (basic Galois theory). The product of the zeros of a monic quadratic is its constant term, so we know that $$ g^6=g\cdot g^5=3. $$ Therefore $g^{12}=3^2=4=-1$, and we can conclude that the order of $g$ is not a factor of $12$. We still have to rule out the possibility that $g$ could be of order $8$. If $g$ were of order $8$, then $g^2$ is of order four. But all the fourth roots of unity are in the prime field $\mathbb{F}_5$. OTOH from the minimal polynomial we read that $g^2=-(2g+3)=3g+2$ is not $\in \mathbb{F}_5$. The claim follows from this.

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Hint: if you actually meant "irreducible polynomial" then

$$x^2+2x+3\in\Bbb Z_5[x]\implies \Delta=b^2-4ac=4-12=-8=2\pmod 5$$

Is there $\,a\in\Bbb Z_5\;\;s.t.\,\,a^2=2\,$ ?

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About monic irreducibles, the reducible ones are those which can be factored in $\mathbb{Z}_5$, i.e., can be written $(x - a) (x - b)$. This means selecting 2 out of 5 as roots if different, and 5 if equal, for a total of: $$ \binom{5}{2} + 5 = 10 + 5 = 15 $$ The total number of monic polynomials $x^2 + c_1 x + c_2$ is just $5 \cdot 5 = 25$ (can select $c_1$, $c_2$ at will), so there are $25 - 15 = 10$ irreducible ones.

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You are right, I had computed the total number as 15 in my head and wrote down a more detailed derivation mixing up the numbers. Thanks! – vonbrand Apr 11 '13 at 12:28

I came to this idea thanks to Jyrki Lahtonen:

$ x^1 = x\\ x^2 = -2x - 3 = 3x + 2\\ x^4 = (4x + 2)(4x + 2) = 4x + 2\\ x^8 = x^4 \cdot x^4 = 4x + 1\\ x^{12} = x^8 \cdot x^4 = 4\\ x^{24} = x^{12} \cdot x^{12} = 16 = 1\\ $

It answers the first part of the question only but completely.

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