Which of these rings are integral domains?

Which of the following rings are integral domains?
(a) $\{a+b\sqrt{5}:a,b\in \mathbb{Q}\}$
(b) the ring of continuous functions from $[0,1]$
(c) the polynomial ring $\mathbb{Z}[x]$.
(d) the ring of complex analytic functions on the disc $\{z\in\mathbb{C}: |z|<1\}$

I know that (a) and (c) are integral domain. i also see in Wikipedia that (b) is not a integral domain but I cannot find any counter example.
For (d) I have no idea.

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For $(b)$ consider the two functions $$f_1(x)=\begin{cases}0 &:\:0\leq x\leq\frac{1}{2}\\ x-\frac{1}{2} &:\: \frac{1}{2}<x\leq 1\end{cases}\quad\text{and}\quad f_2(x)=\begin{cases}\frac{1}{2}-x &:\: 0\leq x\leq\frac{1}{2}\\ 0&:\: \frac{1}{2}<x\leq1\end{cases}.$$ It is easy to check the functions are continuous, but $(f_1\cdot f_2)(x)=0$ for all $x\in[0,1]$.

For $(d)$, it is indeed an integral domain. Note that a nonzero analytic function has the property that the zeroes do not have an accumulation point on a connected domain. Hence, let $Z_1$ be the set of zeroes for $f_1$ and let $Z_2$ be the set of zeroes for $f_2$. Since neither has an accumulation point, neither will $Z_1\cup Z_2$, i.e., the set of zeroes for $f_1\cdot f_2$ will not contain an accumulation point, hence isn't zero. This implies that if $f_1\cdot f_2(z)=0$, then at least one of the $f_1$ or $f_2$ is $0$.

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For (b):

$$f(x)=\begin{cases}0&\text{if}\;\;\;0\leq x<\frac{1}{2}\\2x-1&\text{if}\;\;\;\frac{1}{2}\leq x\leq 1\end{cases}$$

$$g(x)=\begin{cases}-2x+1&\text{if}\;\;\;0\leq x<\frac{1}{2}\\0&\text{if}\;\;\;\frac{1}{2}\leq x\leq 1\end{cases}$$

For (d):

You may need to know the uniqueness theorem (or one of its consequences): an analytic function on a connected domain cannot have a zero which is an accumulation pint unless it is the zero function, or in other words: the zeros of a non identically zero analytic function on a connected domain are isolated (and if the domain is bounded there are thus a finite number of them).

Can you take it from here and show we have an integral domain here?

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The first of your examples is for (b) I believe because $\Bbb{Z}[x]$ is an integral domain. –  fpqc Dec 28 '12 at 8:01
Of course, as my examples aren't polynomials. Thanks. –  DonAntonio Dec 28 '12 at 11:05