# Math Help - abstract algebra hw problem

1. ## abstract algebra hw problem

Hello,

I can't seem to figure out a part of this question:

Let F be a field and let I = { a(sub n)x^n + a(sub n-1)x^(n-1) + ... + a (sub 0), where the a(sub i) are elements of F and a(sub n) + ... + a (sub 0) = 0}. Show that I is an ideal of F[x].

Here's what I have so far: So I defined an f(x) and g(x) that were elements of I and showed that f(x) - g(x) is in I. I'm trying to use the ideal test so I need to show now that, for an r(x) that's an element of F[x], r(x)f(x) and f(x)r(x) are in I. I can't figure out how to do this.

Any help is much appreciated, thanks!

2. Hi rush,

Let $r \in F \ { } f(x)=a_nx^n +a_{n-1}x^{n-1}+...+a_0 \in I$

then the right ideal $rf =\ { } r(a_nx^n +a_{n-1}x^{n-1}+...+a_0)$ distributing will give us

$(ra_n)x^n +(ra_{n-1})x^{n-1}+...+(ra_0)$

Is this in the ideal? note each of the coefficients $ra_i \in F$ so all we need is for the coefficients to sum to 0
$ra_n +ra_{n-1}+...+ra_0 =r(a_n +a_{n-1}+...+a_0) =r(0)=0$

this places $rf \in I$

The other half I leave up to you

3. doesn't the r really have to be an element in F[x] and not just one in F?

4. Sorry i misunderstood the definition of your F[x]. Now i understand that it is the set of linear combinations of x with elements in the field F. Then,
regardless the element $r(x) \in F[x]$ will be of the form

$r(x)= b_n x^n +b_{n-1} x^{n-1}+...+b_o | b_i \in F$

but applying the same technique as before will be a little messy so an alternative you can try is to use the fact that

$a_n +a_{n-1}+...+a_0 =0 \iff f(1)=0$ Since you are just summing the coefficients of f(x). Also note this is a characteristic of all elements in the ideal I.
Using this result we can multiply the two functions and evaluate their product at 1

$h(x)=r(x)f(x)= (b_n x^n +b_{n-1} x^{n-1}+...+b_o)(a_nx^n +a_{n-1}x^{n-1}+...+a_0)$

$h(1)=r(1)f(1)=(b_n (1)^n +b_{n-1} (1)^{n-1}+...+b_o)(a_n(1)^n +a_{n-1}(1)^{n-1}+...+a_0)$

$(b_n +b_{n-1}+...+b_o)(a_n +a_{n-1}+...+a_0)=(b_n +b_{n-1}+...+b_o)(0)=0$

$h(x)=r(x)f(x) \in I$

Hope this helps

5. yes it does, thank you!