Prove or Disprove $I=\{a_0x^0+\dots +a_nx^n:a_0+\dots+a_n=0\}$ is subring of field F[x]
We need to show it is a subring that is
- closed addition
- closed multiplication (struggling)
- $\exists $ neutral additive identity
- $\exists $ additive inverse
Closed addition
Suppose $f(x),g(x) \in I$ where $$f(x)=a_0x^0+\dots +a_nx^n \text{ s.t } a_0+\dots+a_n=0$$ $$g(x)=b_0x^0+\dots +b_nx^n \text{ s.t. } b_0+\dots+b_n=0$$ Now, we need to add them and chech that they $h(x)=g(x)+f(x) \in I$
So, $$ h(x)=f(x)+g(x)=(a_0+b_0)x^0+\dots +(a_b+b_n)x^n$$ where $a_0+\dots a_n=0$ and $b_0+\dots b_n=0$. So, $$0=0+0=(a_0+\dots a_n)+(b_0+\dots b_n)=(a_0+b_0)+\dots+(a_n+b_n) $$ So the coeffinets of h(x) add to zero.
Closed multiplication (struggling at this point) Existence of $0_{F{x}}$
$0_{F[x]}=0x^0+\dots+0x^n $. It is clear that $0=0+\dots+0$
$Existence of additive inverses$
Now we know that F[x] is a field so there are additive inverses for its constants.where $$f(x)-f(x)=(a_0-a_0)x^0+\dots+(a_n-a_n)x^n=0_F$$ Now,$ a_0+\dots+a_n=0$ so,$$0=(-1)(0)=-1(a_0+\dots+a_n)=-a_0\dots-a_n$$ So the coefficients of $-f(x)$ add up to zero so $-f(x)\in I $
could use some help to show that it is closed under multiplication And to show that the coefficients add up to 0.
If you have two elements of $I$, say $u=a_0+a_1x+...+a_nx^n$ and $v=b_0+b_1x+...+b_mx^m$, then their product is
$uv=(a_0+a_1x+...+a_nx^n)(b_0+b_1x+...+b_mx^m)=c_0+c_1x+...+c_{m+n}x^{m+n}$.
Then, plugging $x=1$ we have:
$c_0+c_1+...+c_{m+n}=(a_0+a_1+...+a_n)(b_0+b_1+...+b_m)=0$ since sum of $a_j$'s and $b_j$'s is zero by initial assumption.
Hence $uv\in I.$