**Question:**

The number of irreducible polynomials of the form \(x^2+ax+b\) , with \(a,b\) in the field \(\mathbb{F}_7\) of 7 elements is:

A. 7

B. 21

C. 35

D. 49

**Discussion:**

First, what is the number of polynomials of the form \(x^2+ax+b\) in \(\mathbb{F}_7\) ? \(a\) has 7 choices and \(b\) has 7 choices. So we have a total of \(7 \times 7 =49\) monic degree 2 polynomials in \(\mathbb{F}_7\).

How many of these are reducible? If a monic polynomial of degree 2 is reducible then it must break into two factors in the form \(p(x)=(x-\alpha)(x-\beta) \).

We want to count the number of polynomials of the form \(p(x)=(x-\alpha)(x-\beta)\). This will give us the number of reducible polynomials and we will subtract it from the total number to get the number of irreducibles.

For the counting purposes, we must be careful not to have repetitions. For \(\alpha=0\) we can allow \(\beta=0,1,..,6\). For \(\alpha=1\) we can allow \(\beta=1,…,6\) (NOT 0 because that will again give \((x-0)(x-0)\) which we have already counted in the \(\alpha=0\) situation ).

For \(\alpha=2\) we have \(\beta=2,…,6\) and so on… for \(\alpha=6\) we have \(\beta=6\) only.

So in total, we have \(7+6+…+1=\frac{7\times 8}{2} = 28 \) polynomials which are reducible, has degree 2 and is monic.

Therefore, the number of irreducible polynomials are \(49-28=21\).

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