# Irreducibilty in F[x]

• Apr 24th 2010, 12:04 PM
empressA88
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?
• Apr 24th 2010, 02:56 PM
NonCommAlg
Quote:

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 $g$ on the interval [0,1] exists such that

$(g(x))^2+x=0,$ for all $x \in [0,1],$ which is trivially true.
• Apr 24th 2010, 03:10 PM
empressA88
$(g(x))^2+x=0,$ for all $x \in [0,1],$

so once i prove this equation, it should cover it?
• Apr 24th 2010, 03:15 PM
tonio
Quote:

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

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

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

Tonio
• Apr 24th 2010, 03:16 PM
NonCommAlg
Quote:

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

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.
• Apr 24th 2010, 03:18 PM
empressA88
ok..thank you