A question about commutative rings and homomorphisms

[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?

Thanks in advance

 

[tiny-tim]

Hi Artusartos!

I'm confused …

isn't 1 = 1 + 0x + 0x² + … + 0xⁿ ? ​

 

[Artusartos]

Hi Artusartos!

I'm confused …

isn't 1 = 1 + 0x + 0x² + … + 0xⁿ ? ​

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...

 

[micromass]

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?

 

[Artusartos]

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?

Thanks. I think I understand it now. :)

 

[tiny-tim]

… the function ε : R[x] → R, defined by
\epsilon : a_0 + a_1x + a_2x +· · ·+a_n x^n \rightarrow a_0 …

Do we have the freedom to choose the coefficients ourselves?

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 ​

 

[Artusartos]

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 ​

Thank you.