# Thread: Quotient Rings in Galois Theory - Help

1. ## Quotient Rings in Galois Theory - Help

Hi

I have the proof of this statement:

'Let f(x) be a member of K[x] where f(x) is non-zero and K is a field.
Suppose that K[x]/K[x].f(x) is a field.
Then f(x) is irreducible.'

The proof begins with:
'Assue f(x) is not irreducible.

If f(x) is a unit implies K[x].f(x) = K[x]'
Is this becasue f(x) is a unit, so it has an inverse so multiplying f(x) by it's inverse multiplied by another element in K[x] can create any element in K[x]? (Due to K[x] being closed under multiplication)
ie: a(x)b(x)g(x) = a(x) where b(x) = inverse f(x)
a(x) a member of K[x]

The next bit is that the previous line implies
' K[x]/K[x].f(x) = {0'} the equivalence class of zero'

This is saying that every element is equivalent to zero. Is this because if we have:
a' = b' (the equivalence classes)
then a(x)-b(x) is a member of K[x].f(x) = K[x]
This is true for all pairs of elements a(x) and b(x) in K[x] so all are equivalent to 0.
ie: a(x) - 0 is a member of K[x]
so a' = 0' for all a'

I know there is a lot of info here, but it's been bothering me for ages. I can't get my head around quotient rings.

2. Originally Posted by alanaj5
Hi

I have the proof of this statement:

'Let f(x) be a member of K[x] where f(x) is non-zero and K is a field.
Suppose that K[x]/K[x].f(x) is a field.
Then f(x) is irreducible.'
If $f$ was reducible then we can write $f=gh$ where $\deg g,\deg h < \deg f$. Then $[g],[h]\not = [0]$ since $f\not | g,f\not |h$ but $[gh] = [f] = [0]$. Thus, $K[x]/K[x](f)$ is not an integral domain and so cannot be a field.

3. Originally Posted by ThePerfectHacker
If $f$ was reducible then we can write $f=gh$ where $\deg g,\deg h < \deg f$. Then $[g],[h]\not = [0]$ since $f\not | g,f\not |h$ but $[gh] = [f] = [0]$. Thus, $K[x]/K[x](f)$ is not an integral domain and so cannot be a field.

I get why f=gh and that neither g or h is a unit.
But I don't understand why we know $f\not | g,f\not |h$ and why this gives that $[g],[h]\not = [0]$ and why $[gh] = [f] = [0]$

4. Originally Posted by alanaj5
I get why f=gh and that neither g or h is a unit.
But I don't understand why we know $f\not | g,f\not |h$ and why this gives that $[g],[h]\not = [0]$ and why $[gh] = [f] = [0]$
Remember what $[g]$ means it is the set $\{ g + fq |q\in K[x] \}$. Thus, if $g_1\equiv g_2(\bmod f)$ then $[g_1] = [g_2]$. We cannot have that $f|g$ because the degree of $g$ is less than the degree of $f$. Therefore, $f\not |g$ and so it means $[g]\not =[0]$ because otherwise if $[g]=[0]$ then $g\equiv 0(\bmod f) \implies f|g$ - a contradiction.

5. Thanks, I understand now. I hadn't thought about the quotient ring as modf before. It makes it so much easier.