Ideal of a ring polynomial?

Question

I'm doing a past exam paper and this is the question I am stuck on.

Define the conditions on a subset $I$ of a ring $R$ to be an ideal of $R$. Show that the set $I = \{f ∈ \mathbb{R}[X]|f(1) = 0\}$ is an ideal of $\mathbb{R}[X]$.

I know the conditions for I to be an ideal of R, but I'm confused about what the set actually means? What is the result of $f(a+b)$ if $a,b∈ \mathbb{R}$

Thanks in advance for any help!

Answer 1

$I$ is a subset of $\mathbb{R}[X]$, the ring of polynomials with real coefficients. Each polynomial can be viewed as a function on the real numbers; the set $I$ consists of particularly those functions which evaluate to $0$ at $x=1$.

The ring structure on $\mathbb{R}[X]$: You can add elements in $\mathbb{R}[X]$ as follows: given polynomials $f,g$, $f+g$ is the polynomial defined by the rule $(f+g)(x) = f(x) + g(x)$. Similarly, to multiply polynomials, $fg$ is the polynomial defined by $(fg)(x) = f(x)g(x)$.

Answer 2

The set is a collection of elements. :) $f(a+b)$ is a number, but it's not really important here because You have not to do this operation. You rather need $f(x)+g(x)$.

Answer 3

The set $ I $ defined by you, in fact, is an ideal generated by $x-1$. That is ,any element of $ I$ is of the form $(x-1)g(x)$ for arbitrary $g(x)\in \mathbb R[x]$. The meaning of $f(a+b)$ is nothing just a valuation of $f(x)$ at $x=a+b$.