Questions regarding Rings. I barely passed abstract algebra when I was in college, and 3 years later I bought a book and studied on my own.  And currently I am having trouble with Rings with certain conditions.

Let $\mathbb R$ be the field of real numbers and $\mathbb R[x]$ the ring of polynomials in $x$ with coefficients in $\mathbb R$.
Condition: The subset $S$ is all polynomials whose degree is an even integer, together with the zero polynomial.
I would like to show that this $S$ is not a subring.
My understanding is that to show that a set with two binary operations $+$ and $\cdot$ is a ring iff:
(a) $\langle S,+\rangle$ is an abelian group;
(b) $\langle S,\cdot\rangle$ is a semigroup. ($\cdot$ is associative);
(c) The distributive law holds.
I could be spending few minutes checking every single condition, but this is a problem from the GRE and I cannot do that.
I think it satisfies (b) because multiplying even degree polynomials results in another even degree polynomial.
I don't think (a) is a problem because the degree of the polynomials shouldn't seem to affect if it is an abelian additive group...
Is it (c) ? or am I missing something?

Another question about a ring $T$ such that $s = s^2, \forall s \in T$.
I want to argue that since $s^2-s = 0 \Rightarrow s = 0 \text{,} 1$, so this is the ring $\mathbb Z_2$.
But my instinct tells me that this is not true.
The problem asks me:

Which of the following are true:
I. $s+s=0,\forall s \in T$
II. $(s+t)^2=s^2+t^2, \text{for each} s,t\in T$
III. $T$ is commutative.

I guessed yes to everything according to my assumption that $T \equiv \mathbb Z_2$, only to be disappointed.
 A: We have $x^2 \in S$ and also $x^2 + x \in S$. But...

As for your second problem, you have established that for all $s \in T$:
$$s(s-1) = 0$$
and you have assumed that this implies $s = 0$ or $s-1 =0$. This suggests that you read up the definition of a zero divisor. General rings (especially non-commutative ones) can behave in "strange" ways.
A: Every subset of an algebra which is closed under its operations has the same signature - is a subalgebra of the same type. Because $\Bbb{R}$ is an integral domain, $\Bbb{R}[x]$ is, too. This insures the sum of degrees of two polynomials is the degree of their product, so the algebra is indeed closed under multiplication. For it to be closed under addition, there must be no elements $r$ of the ground ring ($\Bbb{R}$) such that $r+...+r=0,$ as by distributivity this means $x^2r+...+x^2r=0.$ Thus one can add $x^2r$ to $x^2+x$ some number of times to get $x$. $\Bbb{R}$ is of characteristic 0, and $\Bbb{R}[x]$ is freely generated over it, so this is not a problem. But subtraction is also possible, and $x^2$ can be subtracted from $x^2+x$ to get a degree-0 polynomial. However, restricted to positive leading coefficients, it would be closed as a rig - a ring without negative elements, or a monoid that distributes over a commutative monoid.
