1. ## 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?

2. 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.

3. I see!
So my method was wrong, but after rechecking, it seems 2 and 3 are indeed the correct values.

4. 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.