# Commutative Ring, Subgroups, Subrings, Ideals

• May 6th 2009, 03:28 AM
Jimmy_W
Commutative Ring, Subgroups, Subrings, Ideals
Let $\mathcal{F}$ be the set of functions $f:\mathbb{R} \ \rightarrow \ \mathbb{R}$. Given that $\mathcal{F}$ forms a commutative ring with indentity under (pointwise) addition and multiplication,

(a) Find a subgroup of $\mathcal{F}$ which is not a subring. Why?
(b) Find a subring of $\mathcal{F}$ which is not an ideal. Why?
(c) Find an ideal of $\mathcal{F}$ which is not the zero ring or $\mathcal{F}$. Can this ideal be written as a principal idea? Why?
• May 6th 2009, 03:56 AM
NonCommAlg
i'll leave the "whys" for you to figure out!

Quote:

Originally Posted by Jimmy_W
Let $\mathcal{F}$ be the set of functions $f:\mathbb{R} \ \rightarrow \ \mathbb{R}$. Given that $\mathcal{F}$ forms a commutative ring with indentity under (pointwise) addition and multiplication,

(a) Find a subgroup of $\mathcal{F}$ which is not a subring. Why?

set of all polynomials with coefficients in $\mathbb{R}$ and of degree at most 1. this is an additive subgroup which is not closed under multiplication.

Quote:

(b) Find a subring of $\mathcal{F}$ which is not an ideal. Why?

set of all constant functions or, if you want the subring to have the identity element of $\mathcal{F},$ then $\mathbb{R}[x]$ would be an example.

Quote:

(c) Find an ideal of $\mathcal{F}$ which is not the zero ring or $\mathcal{F}$. Can this ideal be written as a principal idea? Why?

fix an $a \in \mathbb{R}$ and define $I_a=\{f \in \mathcal{F}: \ f(a)=0 \}.$ this is a principal ideal. why?