# Values that make a field (need answer checked)

• Dec 2nd 2009, 10:32 AM
elninio
Values that make a field (need answer checked)
I just need my answer checked for this particular problem.

The question is:
"For which values of a=1,2,3,4 is \$\displaystyle Z_5[x]/<x^2+a>\$ a field? Show your work.

We need to know when the polynomial is irreducible in Z_5.
For this, we check to see if a is a perfect square in Z_5. The squares are:
0=0 1=1 2=4 3=4 4=1
So the values of a which arnt squares are 2 and 3, these being the values that give a field.

Correct?
• Dec 2nd 2009, 10:47 AM
Haven
Since \$\displaystyle x^2 +a\$ is a quadratic polynomial, \$\displaystyle x^2 +a\$ either has two linear factors or it's irreducible. If a polynomial \$\displaystyle f(x)\$ has a linear factor \$\displaystyle (x-b)\$ then \$\displaystyle f(b) = 0.\$

So we check for every value of a, if and when \$\displaystyle x^2 + a = 0\$

e.g. \$\displaystyle f(x) = x^2 + 1 \$
\$\displaystyle f(0) = 1, f(1) = 2, f(2) = 5 = 0\$. So \$\displaystyle f(x)\$ is not irreducible.
• Dec 2nd 2009, 10:52 AM
elninio
I see!
So my method was wrong, but after rechecking, it seems 2 and 3 are indeed the correct values.
• Dec 2nd 2009, 11:00 AM
Bruno J.
Your method was correct. You are both essentially doing the same thing - eliminating the values of \$\displaystyle a\$ for which the polynomial has a root.