1. ## Irreducibilty in F[x]

Let f:[0,1]→R be defined by f(x)=x, show that the polynomial x^2+f in T[x] is irreducible, where T is the ring of continuous real valued function on [0,1].

For irreduciblity do i just show that this does not have any roots. by evaluating the equation x^2 + f, at 0 and 1?

2. Originally Posted by empressA88
Let f:[0,1]→R be defined by f(x)=x, show that the polynomial x^2+f in T[x] is irreducible, where T is the ring of continuous real valued function on [0,1].

For irreduciblity do i just show that this does not have any roots. by evaluating the equation x^2 + f, at 0 and 1?
the question is kind of strange because the continiuity condition is useless here. you need to prove that no real-valued (continuous) function $\displaystyle g$ on the interval [0,1] exists such that

$\displaystyle (g(x))^2+x=0,$ for all $\displaystyle x \in [0,1],$ which is trivially true.

3. $\displaystyle (g(x))^2+x=0,$ for all $\displaystyle x \in [0,1],$

so once i prove this equation, it should cover it?

4. Originally Posted by empressA88
$\displaystyle (g(x))^2+x=0,$ for all $\displaystyle x \in [0,1],$

so once i prove this equation, it should cover it?

Of course: since the polynomial $\displaystyle x^2+f$ has degree $\displaystyle \leq 3$ , it reducible iff it has a root in $\displaystyle T\iff \,\,\exists g\in T$ s.t. $\displaystyle g^2+f=0\in T[x]\iff g^2(t)+t=0\,\,\,\forall\,t\in [0,1]$ , since the zero element in the ring $\displaystyle T$ is the zero map...and this is what NCA told you is trivially true.

Tonio

5. Originally Posted by empressA88
$\displaystyle (g(x))^2+x=0,$ for all $\displaystyle x \in [0,1],$

so once i prove this equation, it should cover it?
yes, and there's nothing to prove! just put x = 1, or any positive number, to see that the equality cannot hold.

6. ok..thank you