# Quotient ring and field question

• Apr 17th 2010, 06:07 AM
gizmo
Quotient ring and field question
For which prime numbers p is the quotient ring $\displaystyle F_5[x]/(x^2+1)$ a field?

From theorems in my notes, I know this quotient ring is a field if $\displaystyle (x^2+1)$ is a maximal ideal, i.e $\displaystyle (x^2+1)$ is irreducible modulo p over the field $\displaystyle F_5[x]$.

I don't see where to go from here though.. any help would be great :)
Thank you!
• Apr 17th 2010, 06:59 AM
tonio
Quote:

Originally Posted by gizmo
For which prime numbers p is the quotient ring $\displaystyle F_5[x]/(x^2+1)$ a field?

From theorems in my notes, I know this quotient ring is a field if $\displaystyle (x^2+1)$ is a maximal ideal, i.e $\displaystyle (x^2+1)$ is irreducible modulo p over the field $\displaystyle F_5[x]$.

I don't see where to go from here though.. any help would be great :)
Thank you!

Twice you wrote $\displaystyle \mathbb{F}_5[x]$ when it should, imo, be $\displaystyle \mathbb{F}_p[x]$

Hint: -1 is a quadratic residue modulo p iff $\displaystyle p=1\!\!\!\mod 4$

Tonio

Ps. By the way, precisely in $\displaystyle \mathbb{F}_5[x] \,,\,\,(x^2+1)$ is a not a maximal ideal. (Happy)
• Apr 18th 2010, 02:58 AM
gizmo
oops sorry! yes I do mean $\displaystyle F_p[x]$

so, $\displaystyle x^2=-1modp$ iff $\displaystyle p=1mod4$

which is saying $\displaystyle (x^2+1)$ is reducible iff $\displaystyle p=1mod4$

but we want to find p when $\displaystyle (x^2+1)$ is irreducible:

$\displaystyle x^2 \neq -1modp$ or $\displaystyle p \neq 1mod4$

I'm not really sure what to do from here though :/
• Apr 18th 2010, 08:11 AM
tonio
Quote:

Originally Posted by gizmo
oops sorry! yes I do mean $\displaystyle F_p[x]$

so, $\displaystyle x^2=-1modp$ iff $\displaystyle p=1mod4$

which is saying $\displaystyle (x^2+1)$ is reducible iff $\displaystyle p=1mod4$

but we want to find p when $\displaystyle (x^2+1)$ is irreducible:

$\displaystyle x^2 \neq -1modp$ or $\displaystyle p \neq 1mod4$

I'm not really sure what to do from here though :/

You have it all above! First, a pol. of degree $\displaystyle \leq 3$ is reducible (over a given field) iff it has a root there; next, an ideal $\displaystyle <f(x)> \leq \mathbb{F}[x]\,,\,\,\mathbb{F}$ a fiedl, is maximal iff $\displaystyle f(x)$ is irreducible.

So now you know exactly over what fields $\displaystyle \mathbb{F}_p$ the polynomila $\displaystyle x^2+1$ is irreducible , so...

Tonio
• Apr 18th 2010, 10:42 AM
gizmo
so if p $\displaystyle \neq$ mod4 ... p must satisfy one of these:

p=0mod4 but this is not true as a prime number can't be divided by 4.
p=2mod4 but this is also not true as it would mean p is even.

Therefore, p=3mod4, in order for the quotient ring to be a field.

I think that's it :)
Thank you for your help! I would rep you if I could again :p