# A question about commutative rings and homomorphisms

1. Jan 6, 2013

### Artusartos

Let R be a commutative ring. Show that the function ε : R[x] → R, defined by
$$\epsilon : a_0 + a_1x + a_2x +· · ·+a_n x^n \rightarrow a_0$$,
is a homomorphism. Describe ker ε in terms of roots of polynomials.

In order to show that it is a homomorphism, I need to show that ε(1)=1, right?

But $$\epsilon(1) = a_0+a_1+...+a_n \not= 1$$

So can anybody help me with this?

2. Jan 6, 2013

### tiny-tim

Hi Artusartos!

I'm confused

isn't 1 = 1 + 0x + 0x2 + … + 0xn ?

3. Jan 6, 2013

### Artusartos

Oh...I thought we had to do it the other way around...

But why do we set $$1=a_0 + a_1x + ... + a_nx^n$$ and then choose a_0 to be 1 and the rest of the coefficients to be zero? Do we have the freedom to choose the coefficients ourselves?

I thought we were supposed to compute $$\epsilon(1)$$ and then find the multiplicative identity of $$a_0$$, and show that their equal? But the multiplicative identity for $$a_0$$ is 1 and $$\epsilon(1) = a_0 + a_1 + ... +a_n \not = 1$$. I must be doing something wrong...

4. Jan 6, 2013

### micromass

Staff Emeritus
You seem to misunderstand the mapping $\varepsilon$. The mapping is

$$\varepsilon( a_0 + a_1X+...+a_nX^n)=a_0$$

For example:

$$\varepsilon ( 2 + 3X + 4X^2)=2$$

because $a_0=2,~a_1=3,~a_2=4$.

Other examples:

$$\varepsilon ( 3 + 321X)=3$$
$$\varepsilon (421)=421$$
$$\varepsilon ( 1 + X^2 + 5X^4+6X^{77})=1$$

Does that clear things up for you?

5. Jan 6, 2013

### Artusartos

Thanks. I think I understand it now. :)

6. Jan 6, 2013

### tiny-tim

ah, the a's aren't constants in the function ε,

they're variables (coordinates) in R …

ε sends the variable point a0 + a11x + a2x +· · ·+a_n xn to the point a0

7. Jan 6, 2013

Thank you.