Q. \(x^4 + x + 2\) irreducible over \(\text{GF}(3)\).

Problem:

Show that the polynomial \(f(x)=x^4+x+2\) is irreducible over the finite field \(\mathrm{GF}(3)\) (also known as \(\mathbb{Z}_3\)).

1. Step 1 – Understand the criteria for irreducibility:

   A polynomial of degree \(4\) is irreducible if and only if it has no roots in the field (no linear factors) and cannot be factored into the product of two irreducible quadratic polynomials.

2. Step 2 – Check for roots in \(\mathrm{GF}(3)\):

   The elements of \(\mathrm{GF}(3)\) are \(\{0,1,2\}\). We evaluate \(f(x)\) at each element:

   • For \(x=0\): \(f(0)=0^4+0+2=2 \equiv 2\pmod{3}\).
   • For \(x=1\): \(f(1)=1^4+1+2=4 \equiv 1\pmod{3}\).
   • For \(x=2\): \(f(2)=2^4+2+2=16+2+2=20 \equiv 2\pmod{3}\).

   Since \(f(x) \neq 0\) for any \(x\) in \(\mathrm{GF}(3)\), the polynomial has no linear factors.

3. Step 3 – Assume a factorization into two quadratics:

   Since there are no linear factors, if \(f(x)\) is reducible, it must be the product of two monic quadratic polynomials. We set up the identity:

   \[ x^4+x+2=(x^2+ax+b)(x^2+cx+d) \]

   Expanding the right side gives:

   \[ x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd \]

4. Step 4 – Solve the system of equations over \(\mathrm{GF}(3)\):

   Comparing coefficients, we get:

   1. \(a+c=0 \implies c=-a\)
   2. \(ac+b+d=0 \implies -a^2+b+d=0 \implies b+d=a^2\)
   3. \(ad+bc=1 \implies ad-ab=1 \implies a(d-b)=1\)
   4. \(bd=2\)
5. Step 5 – Test values for \(a\):

   From equation (iii), \(a\) cannot be \(0\) because \(0 \neq 1\). Therefore, \(a\) must be \(1\) or \(2\).

   Case 1: \(a=1\)

   • From (ii): \(b+d=1^2=1\).
   • From (iii): \(1(d-b)=1 \implies d-b=1\).
   • Adding these: \(2d=2 \implies d=1\). Then \(b=0\).
   • Check (iv): \(bd=0 \cdot 1=0\). But we need \(bd=2\). Contradiction.

   Case 2: \(a=2\)

   • From (ii): \(b+d=2^2=4 \equiv 1\pmod{3}\).
   • From (iii): \(2(d-b)=1 \implies d-b=2\) (since \(2 \cdot 2=4 \equiv 1\)).
   • Adding these: \(2d=3 \equiv 0 \implies d=0\). Then \(b=1\).
   • Check (iv): \(bd=1 \cdot 0=0\). But we need \(bd=2\). Contradiction.
6. Step 6 – Conclusion:

   Since there are no possible values for the coefficients \(a,b,c,d\) in \(\mathrm{GF}(3)\) that satisfy the factorization, \(f(x)\) cannot be decomposed into lower-degree polynomials. Therefore, \(x^4+x+2\) is irreducible over \(\mathrm{GF}(3)\).