Quotient rings and homorphic images

[simo1]

am given that ϕ is a function from F(R) tp RxR defined by ϕ(f)=(f(0),f(1))
i proved that ϕ is a homomorphism from F(R) onto RxR.
i showed that
1) ϕ(f) +ϕ(g)=ϕ(f+g) [for all f,g in F(R)]
2)ϕ(f)*ϕ(g)= ϕ(f*g)

how do i show that ϕ is onto and define the kernal??(Wasntme)

 

[Deveno]

Re: quotient rings and homorphic images

To show that $\phi$ is onto, given $(a,b) \in \Bbb R \times \Bbb R$ you have to exhibit one (there may be more than one) $f \in F(\Bbb R)$ with $f(0) = a, f(1) = b$.

You haven't said what kind of function-space $F(\Bbb R)$ is: in fact, I am only guessing that by R you mean the real numbers.

I suspect that the function $f(x) = (b-a)x + a$ will do the trick, but in order for us to help you, we need YOU to define your terms. We all aren't using YOUR textbook, or lecture notes.

The kernel of $\phi$ is all $f \in F(\Bbb R)$ such that $f(0) = f(1) = 0$. That's about as specific as I can get without more information.

 

[simo1]

Re: quotient rings and homorphic images

To show that $\phi$ is onto, given $(a,b) \in \Bbb R \times \Bbb R$ you have to exhibit one (there may be more than one) $f \in F(\Bbb R)$ with $f(0) = a, f(1) = b$.

You haven't said what kind of function-space $F(\Bbb R)$ is: in fact, I am only guessing that by R you mean the real numbers.

I suspect that the function $f(x) = (b-a)x + a$ will do the trick, but in order for us to help you, we need YOU to define your terms. We all aren't using YOUR textbook, or lecture notes.

The kernel of $\phi$ is all $f \in F(\Bbb R)$ such that $f(0) = f(1) = 0$. That's about as specific as I can get without more information.

yes by R it is real numbers. and everythin that I wrote is as it is on the texbook. are there options here on mathshelpboard that we can use to show real numbers, complex etc

 

[Deveno]

Re: quotient rings and homorphic images

If you use the buttons, there is a button labeled $\Sigma$ that creates \(\displaystyle tags that wrap around your text.

This let's you type expressions in Latex, which makes "pretty math symbols". For example putting:

\sqrt{x^2 + 1}

inside the tags produces:

\(\displaystyle \sqrt{x^2 + 1}\)

You can also enter Latex directly, by using the dollar sign as a delimiter (for in-line latex) or a double dollar sign (which produces "centered display style").

The Latex code for the "blackboard bold" font that is used for the sets you listed is either:

\mathbb{text goes here}

or the code \Bbb, for example:

\Bbb Z

inside \(\displaystyle tags or dollar signs produces:

$\Bbb Z$

**********

You still haven't told us what is meant by $F(\Bbb R)$, although I guess it is a set of functions $f:\Bbb R \to \Bbb R$. Knowing which functions would be nice.\)\)

 

[blue_raver22]

I would first clarify that F(R) is the set of all real-valued functions and RxR is the set of ordered pairs of real numbers. The function ϕ maps each function in F(R) to an ordered pair in RxR, specifically the values of the function at 0 and 1.

To show that ϕ is onto, we need to show that for any ordered pair (a,b) in RxR, there exists a function f in F(R) such that ϕ(f)=(a,b). In other words, we need to show that for any (a,b), there exists a function f such that f(0)=a and f(1)=b.

This can be achieved by considering the constant functions f(x)=a and g(x)=b. Both of these functions are in F(R) and ϕ(f)=(a,a) and ϕ(g)=(b,b), which means that ϕ is onto.

The kernel of ϕ is the set of all functions in F(R) that map to the zero element in RxR, which is the ordered pair (0,0). In other words, the kernel is the set of all functions f such that f(0)=0 and f(1)=0.

To define the kernel, we can rewrite the condition as f(0)-0=0 and f(1)-0=0. This can be interpreted as the set of all functions whose value at 0 and 1 is equal to 0. Therefore, the kernel of ϕ is the set of all constant functions f(x)=0.

In summary, ϕ is a homomorphism from F(R) onto RxR because it preserves addition and multiplication, and it is onto because for any ordered pair (a,b) in RxR, there exists a function in F(R) that maps to it. The kernel of ϕ is the set of all constant functions f(x)=0.