Showing a subset of the functions $\mathbb R \to \mathbb R$ is a subring

Question

Let $T$ be the set of all functions from $\mathbb R$ to $\mathbb R$. Let $S = \{ f \in T : f(2) = 0 \}$. Show that $S$ is a subring of $T$.

$T$ has a zero element and an identity element by inspection.

I am not sure how to work with set of all functions? There are a lot of axioms I need to verify, just looking for a helpful start.

Do you know what a subring is? Do you know what you need to verify? Do you understand the proof that $T$ is a ring? — Ben Millwood, Feb 20 '13 at 20:50
In the future, try to make sure your problem is self-contained inside the question and doesn't rely on the title. Also your use of LaTeX via mathjax is quite strange, refer to the reference for more guidance. — JSchlather, Feb 20 '13 at 20:51
Note that often, when working with rings with $1$, one requires that a subring contain that $1$, which your set does not. — Tobias Kildetoft, Feb 20 '13 at 20:51

score 1 · Answer 1 · answered Feb 23 '13 at 00:39

If by definition one requires subrings of a unitary ring to contain the multiplicative unit of the ring, then your set is not a subring (why?).

Your set though is an ideal and, in fact, a maximal one, since

$$S=\ker\phi\;\;,\;\;\phi:T\to\Bbb R\,\,\,,\,\,\phi(f):=f(2)$$

Now just check $\,\phi\,$ is onto, apply the first isomorphism theorem and also the second "property" here

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