We are considering the ring A of germs at 0 of smooth functions of one variable.
This ring consists of equivalence classes of smooth functions defined in some neighborhood of 0, where two functions are equivalent if they agree on some neighborhood of 0.

We are asked for a prime ideal other than 0, the whole ring A, and the ideal of functions vanishing at 0 (which is the set {f ∈ A | f(0) = 0}).

Recall that in this ring, the ideal of functions vanishing at 0 is prime. But we are to find another one.

One common prime ideal in this ring is the ideal of functions that vanish to order at least 1 at 0. However, note that the ideal of functions vanishing at 0 is already given and we are to avoid that.

Another idea: consider the ideal of functions that vanish to order at least 2 at 0? But note that this is not prime because if we take two functions each vanishing to order 1 (but not higher) then their product might vanish to order 2 and so be in the ideal, but each factor is not in the ideal.

Actually, we need a prime ideal. Let me recall that the ring A is a local ring with maximal ideal m = {f ∈ A | f(0)=0}. The quotient ring A/m is isomorphic to the field of germs of smooth functions at 0 that are constant? Actually, it's isomorphic to the field of real numbers if we are over the reals? Wait, no: the quotient A/m is isomorphic to the real numbers because every germ can be represented by a function and then we evaluate at 0 to get the constant. So A/m ≅ ℝ.

Now, consider the ideal p = {f ∈ A | f(0) = 0 and f'(0) = 0}. This is a subset of m. We claim that this is prime.

Why? Suppose we have two functions g and h such that gh ∈ p. Then we have:
(gh)(0) = g(0)h(0) = 0 => so either g(0)=0 or h(0)=0.
(gh)'(0) = g'(0)h(0) + g(0)h'(0) = 0.

But note that if both g and h are in m (so they vanish at 0) then we have g'(0)h(0)=0 and g(0)h'(0)=0, which doesn't immediately give that either g'(0)=0 or h'(0)=0.

However, we must consider that the ideal p is defined by two conditions: f(0)=0 and f'(0)=0.

So if we have gh in p, then we know:
g(0)h(0) = 0 --> so without loss of generality, assume g(0)=0 (if not, then h(0)=0 and then we are done for the second condition? Not exactly, because we have to check the derivative condition).

Actually, let me write:

Case 1: g(0) ≠ 0. Then from g(0)h(0)=0 we get h(0)=0. Then we need to check the derivative condition for gh to be in p: we have (gh)'(0)= g'(0)h(0) + g(0)h'(0) = g'(0)*0 + g(0)h'(0) = g(0)h'(0). Since g(0)≠0, we have g(0)h'(0)=0 implies h'(0)=0. Then h satisfies both h(0)=0 and h'(0)=0, so h ∈ p.

Case 2: h(0) ≠ 0. Similarly, we get g(0)=0 and then from (gh)'(0)= g'(0)h(0) + g(0)h'(0) = g'(0)h(0) = 0. Since h(0)≠0, we get g'(0)=0. Then g(0)=0 and g'(0)=0 so g ∈ p.

But what if both g(0)=0 and h(0)=0? Then we have (gh)'(0)= g'(0)h(0) + g(0)h'(0) = 0. But we also require that (gh)'(0)=0? Actually, we have that gh is in p so (gh)'(0)=0. But note that we have no condition on g'(0) or h'(0) individually. So it is possible that g and h are both in m (so vanish at 0) but neither has derivative 0, but their product has derivative 0? For example, take g(x)=x and h(x)=x. Then gh(x)=x^2. Then (gh)'(x)=2x, so at 0 it is 0. So gh is in p (since it vanishes at 0 and derivative is 0). But g(x)=x and h(x)=x are not in p because each has derivative 1 (non-zero). So this shows that if we only have gh in p and both g and h vanish at 0, we cannot conclude that either g or h is in p.

Therefore, p as defined above is not prime.

We need a different idea.

Another approach: consider the ideal of functions that vanish to order at least k for a fixed k? But these are not prime for k>1.

Alternatively, we can consider the ideal generated by x^2 (if we take a representative). But note that the ring is germs, so we can take a representative and then consider the ideal generated by that representative. However, the ideal generated by x^2 in A is the set of all functions that are multiples of x^2 (in the germ sense). This is not prime because x is not in the ideal but x * x = x^2 is in the ideal, and x is not in the ideal.

How about the ideal generated by x? That is the set of functions that are multiples of x. But note that this is exactly the set of functions that vanish at 0? Because if f is a multiple of x, then f(0)=0. Conversely, if f vanishes at 0, then we can write f(x) = x * g(x) for some smooth function g (by the division algorithm for smooth functions? Actually, we can do this locally: in a neighborhood, we can define g(x)=f(x)/x for x≠0 and then define g(0)=f'(0) to make it continuous and then smooth? But note that f being smooth and vanishing at 0 implies that f(x)=x*g(x) for some smooth g? This is a standard result: the function g is smooth at 0 if we set g(0)=f'(0). So the ideal generated by x is exactly m.

So we need a different prime ideal.

Let me recall that the ring A is a directed union of rings of smooth functions on intervals [-ε, ε] (or (0,ε) and (-ε,0] but we need to be careful). Also, it is a local ring with maximal ideal m.

There is a prime ideal called the "nilpotent ideal" but in this ring there are no nilpotents? Actually, no: the ideal we are about to describe is not nilpotent.

Another idea: consider the ideal of functions that are flat at 0, i.e., all derivatives at 0 are 0. But note that this ideal is not prime: for example, take f(x)=exp(-1/x^2) for x>0 and 0 for x<=0. Then f is flat at 0. But also take g(x)=exp(-1/x^2) for x>0 and 0 for x<=0. Then fg is exp(-2/x^2) which is also flat. But f and g are not flat? Actually, they are flat at 0? Because all derivatives of exp(-1/x^2) at 0 are 0. However, consider h(x)=x. Then h is not flat. But note that fh = x * exp(-1/x^2). This function is also flat at 0? Actually, no: because the derivative of x * exp(-1/x^2) at 0 is not defined? Wait, let's compute:

h(x) = x * exp(-1/x^2) for x≠0, and 0 at x=0.

h'(x) = exp(-1/x^2) + x * exp(-1/x^2) * (2/x^3) = exp(-1/x^2) (1 + 2/x^2).

As x->0, this goes to 0? Actually, the term 2/x^2 * exp(-1/x^2) is problematic. But note that exp(-1/x^2) decays faster than any polynomial. So the limit of h'(x) as x->0 is 0. But is h' continuous at 0? We have to define h'(0)=0. Then we can check the second derivative and so on. In fact, it is known that h is smooth and all derivatives at 0 are 0. So h is flat. Therefore, the ideal of flat functions is not prime because it contains products of flat functions? But also, if we take two flat functions their product is flat. However, if we take a flat function and a non-flat function, it might be flat? Actually, in this example, we multiplied a flat function (exp(-1/x^2)) by x and got a flat function. So it doesn't help.

How about the ideal of functions that are flat at 0? We claim that this ideal is not prime. Why? Consider the function f(x) = exp(-1/x^2) for x>0 and 0 for x<=0. This is flat. Now consider g(x) = exp(-1/x^2) for x<0 and 0 for x>=0. Then f and g are flat. But their product is 0 for x<=0 and exp(-2/x^2) for x>0. This function is not flat: because at 0, it has a cusp? Actually, the function h(x)=f(x)*g(x) = exp(-2/x^2) for x≠0 and 0 at 0. This function is flat? Yes, because it is the same as exp(-1/(x^2/2)) scaled? Actually, it is known that exp(-1/x^2) is flat and so is exp(-2/x^2). So the product of two flat functions is flat. So that doesn't help.

Alternatively, consider the function k(x) = exp(-1/x^2) for x>0 and exp(-1/x^2) for x<0? Then k is flat? Actually, it is even. But the point is that we need a function that is not flat. How about taking a function that is not flat and break it into two parts? For example, consider:

p(x) = exp(-1/x^2) for x>0 and 0 for x<=0. (flat at 0)
q(x) = 0 for x>=0 and exp(-1/x^2) for x<0. (flat at 0)

Then p and q are both flat. But now consider a function like:

r(x) = exp(-1/x^2) for x>0 and 0 for x<=0. (which is p)
and s(x) = 0 for x>=0 and exp(-1/x^2) for x<0. (which is q)

Then r and s are flat. But what if we take a function that is not flat? For example, the function t(x)=x. Then t is not flat. But note that t is not in the ideal of flat functions? And we are not multiplying two functions that are not in the ideal.

We want to show that if we have two functions f and g that are not flat (so they are not in the ideal of flat functions) then their product might be flat? That would show the ideal is not prime. But it's not clear.

Actually, the ideal of flat functions is not prime. Here's a counterexample: consider the function u(x) = exp(-1/x^2) for x≠0 and 0 at 0. This is flat. Now consider v(x)=x. Then v is not flat. Now, the product uv = x * exp(-1/x^2). As we saw above, this is flat. So we have uv is flat (so in the ideal) but u is flat (so in the ideal) and v is not in the ideal. This does not violate the prime property because for the ideal to be prime we require that if a product is in the ideal then at least one factor is in the ideal. In this case, the product is in the ideal and one factor (u) is in the ideal. So it doesn't help.

We need two functions that are not in the ideal (not flat) but their product is flat. For example, take:

f(x) = exp(-1/|x|) ??? But this is not smooth at 0.

How about:

f(x) = exp(-1/x^2) for x>0 and 0 for x<=0. (flat at 0, so in the ideal)
g(x) = exp(-1/x^2) for x<0 and 0 for x>=0. (flat at 0, so in the ideal)

Then f and g are both in the ideal. Their product is in the ideal. But we want a function not in the ideal.

How about:

f(x) = x for x>=0 and 0 for x<0. This function is not flat: because its derivative at 0 from the right is 1 and from the left is 0, so it's not differentiable at 0? Actually, it is differentiable? Let's check:

f(x) = 0 for x<0, x for x>=0.

f'(x) = 0 for x<0, 1 for x>0. At 0, the derivative from the left is 0, from the right is 1. So it is not differentiable at 0. Therefore, it is not smooth. So we cannot use it.

We need smooth functions. How about:

f(x) = x^2 for x>=0 and 0 for x<0? This is smooth? Check:

f(x) = 0 for x<0, x^2 for x>=0.

f'(x) = 0 for x<0, 2x for x>0. At 0, the derivative from the right is 0, from the left is 0. So f'(0)=0.

f''(x) = 0 for x<0, 2 for x>0. At 0, the second derivative from the right is 2, from the left is 0. So not twice differentiable. So not smooth.

How about the standard bump function that is not flat? For example, take a function that is 0 on [-1,0] and on (0,1] it is a smooth function that is not flat. For example, take a function that is 0 on [-1,0], and on [0,1] it is a smooth function that is not identically zero and has all derivatives at 0 equal to 0? But then it would be flat at 0. We want a function that is not flat.

Actually, we can take a function that is even and non-flat. For example, take the function:

f(x) = exp(-1/x^2) for x>0, and we extend it to x<0 by evenness? But then it becomes:

f(x) = exp(-1/x^2) for x≠0, and 0 at 0.

This is even and flat at 0.

How about taking a function that is not flat and supported in a way that it is not even? For example, take:

g(x) = exp(-1/(x^2+x)) ??? This is not standard.

Alternatively, consider the function:

h(x) = exp(-1/x) for x>0 and 0 for x<=0. This is not smooth at 0? Because the derivative from the right does not exist: the derivative as x->0+ is infinite.

We need a smooth function that is not flat. The simplest is:

k(x) = e^{-1/x^2} for x>0, and we can define it for x<0 by k(x)=0. But then it is not smooth at 0? Actually, it is smooth: it is the standard example of a smooth function with all derivatives at 0 equal to 0. So it is flat.

So how to get a non-flat function? We can take a function that is not identically zero in a neighborhood of 0 and that has a non-zero derivative at 0. For example, the function l(x)=x. This is not flat.

But then if we take two such functions, their product is not flat? Because if f and g are smooth and both have a non-zero derivative at 0, then f*g also has a non-zero derivative at 0? Not necessarily: if they are both non-zero at 0 then the derivative of the product is f'(0)g(0)+f(0)g'(0). If f(0)=0 and g(0)=0, then the derivative is 0. But then we are in the ideal m.

We are stuck? Let me think of a different approach.

In the ring of germs of smooth functions, there are many prime ideals. In fact, it is a non-Noetherian ring with many prime ideals.

One construction is to use the fact that the ring is a union of polynomial rings. Specifically, every smooth germ can be written as a power series. But not exactly: there is a theorem that says that every smooth germ has a Taylor series. But the ideal we are about to construct might be in terms of the Taylor series.

Consider the ideal of functions whose Taylor series at 0 has no constant term and no linear term. That is, the functions that vanish to order 2 at 0. But we saw that this is not prime.

How about the ideal of functions whose Taylor series has no constant term, no linear term, and no quadratic term? That is, vanish to order 3. Again, not prime.

But note: the intersection of all the ideals m^k (where m^k is the ideal of functions vanishing to order at least k) is the zero ideal. And these are not prime.

Another idea: consider a specific function and then take a prime ideal containing it. For example, take the function f(x)=x. Then the ideal (x) is m, which we already know is prime. But we want a different one.

Take a function that is not in m, for example, a constant function. But then the ideal generated by a constant function is the whole ring if the constant is non-zero.

Take a function that is in m but not in the square of m. For example, take g(x)=x. Then the ideal (g) is m. But we want a proper ideal that is prime and not equal to m.

How about the ideal generated by x^2? Then we have an ideal that is not prime because x is not in it but x*x=x^2 is in it.

But if we take a prime ideal containing x^2, then it must contain x. So the only prime ideal containing (x^2) is m. So that doesn't give a new one.

We need a prime ideal that does not contain x. But then it must contain a function that is a unit? Not necessarily.

Let me recall that the ring A has a maximal ideal m, and it is a local ring. So every proper ideal is contained in m. So any prime ideal that is not the whole ring is contained in m. Also, it cannot be the zero ideal. So we are looking for a prime ideal between 0 and m.

Now, consider the ideal of functions that are flat at 0 (i.e., all derivatives at 0 are 0). We call this ideal N. We know that N is a prime ideal? Let me try to prove it.

Suppose f*g is in N. We want to show that either f or g is in N. If both f and g are not in N, then at least one of them has a non-zero derivative at 0. But it's not clear that we can get a contradiction.

However, it is a known fact that the ideal of flat functions is prime in the ring of germs of smooth functions. The reason is that the complement is a union of open sets in the topology of the ring. Specifically, the non-flat functions are the union over k of the set of functions that have a non-zero derivative of order k. But this is not open. Actually, the set of functions that are not flat is dense. So this approach may not work.

But there is a theorem: in the ring of germs of smooth functions, the ideal of flat functions is prime. I think this is true. However, we are asked for a prime ideal other than 0, A, and the maximal ideal. The ideal of flat functions is contained in the maximal ideal m, and it is not the whole ring and not zero. So it is a candidate.

But is it prime? Let me try to prove it.

Suppose fg is flat at 0. We wish to show that at least one of f or g is flat at 0. Suppose f is not flat at 0. This means that there is some derivative of f that is non-zero at 0. Without loss of generality, assume that f^{(k)}(0) != 0 for some k. Then in a neighborhood of 0, f is not flat. Now, what about g? We don't know. But note that the product fg has all derivatives at 0 equal to 0. In particular, we have:

(f*g)^{(n)}(0) = 0 for all n.

By the Leibniz rule, we have:

(f*g)^{(n)} = sum_{i=0}^{n} C(n,i) f^{(i)} g^{(n-i)}

So we have that for every n, the sum above is 0 at 0.

Now, if we assume that f is not flat, then let k be the smallest integer such that f^{(k)}(0) != 0. Then we consider the expansion of (f*g)^{(k)} at 0:

(f*g)^{(k)}(0) = C(k,0) f^{(0)}(0) g^{(k)}(0) + C(k,1) f^{(1)}(0) g^{(k-1)}(0) + ... + C(k,k) f^{(k)}(0) g^{(0)}(0)

But note that if i<k then f^{(i)}(0)=0. So the only term that might be non-zero is the last term: C(k,k) f^{(k)}(0) g^{(0)}(0) = f^{(k)}(0) * g(0).

So we have:

f^{(k)}(0) * g(0) = 0.

Since f^{(k)}(0) != 0, we have g(0)=0.

Now, we consider the next derivative. For n=k+1:

(f*g)^{(k+1)}(0) = 0.

The expansion:

sum_{i=0}^{k+1} C(k+1, i) f^{(i)} g^{(k+1-i)}
But note that for i<k, f^{(i)}(0)=0. For i=k, we have:

C(k+1, k) f^{(k)}(0) g^{(1)}(0) = (k+1) * f^{(k)}(0) * g'(0)

For i=k+1, we have:

C(k+1, k+1) f^{(k+1)}(0) g^{(0)}(0) = f^{(k+1)}(0) * g(0)

But we know g(0)=0, so the last term is 0. Also, for i<k, terms are 0. So the only term is the one with i=k:

(k+1) * f^{(k)}(0) * g'(0) = 0.

Since f^{(k)}(0) != 0, we have g'(0)=0.

Similarly, we can continue. For n=k+2:

(f*g)^{(k+2)}(0) = 0.

The expansion:

sum_{i=0}^{k+2} C(k+2, i) f^{(i)} g^{(k+2-i)}

The terms for i<k are 0. For i=k, we have:

C(k+2, k) f^{(k)}(0) g^{(2)}(0) = (k+2)(k+1) f^{(k)}(0) g^{(2)}(0)
For i=k+1, we have:

C(k+2, k+1) f^{(k+1)}(0) g^{(1)}(0) = 2 * f^{(k+1)}(0) * g'(0)

But we know g'(0)=0, so this term is 0. For i=k+2, we have:

f^{(k+2)}(0) * g(0) = 0.

So the only non-zero term is the one with i=k:

(k+2)(k+1) f^{(k)}(0) g^{(2)}(0) = 0.

Hence, g^{(2)}(0)=0.

Continuing this way, we will get that g^{(j)}(0)=0 for all j. Hence, g is flat at 0.

Therefore, if f*g is flat and f is not flat, then g must be flat. This shows that the ideal of flat functions is prime.

But wait, what if there are multiple non-zero derivatives? We assumed that k is the smallest such that f^{(k)}(0)!=0. Then we have that g must have g(0)=0, g'(0)=0, ... up to g^{(k)}(0)=0? Actually, in the above, we only got up to g^{(2)}(0)=0 for n=k+2. But we can go on: for n=k+j, we will get that the term with the lowest derivative of f is the one with i=k, and it will give a term involving g^{(j)}(0). And since the others are zero (because of the earlier conditions on g), we get that g^{(j)}(0)=0. So indeed, g is flat.

Therefore, the ideal N of functions that are flat at 0 is prime.

And it is not the whole ring, not the maximal ideal, and not the zero ideal. Also, it is not the ideal {f: f(0)=0} because there are non-flat functions that vanish at 0 (like x) and are not in N.

So we have found a prime ideal: the ideal of flat functions at 0.

Explicitly, it is:

N = { f ∈ A | f^{(k)}(0) = 0 for all k ≥ 0 }

(Note: for a smooth function, being flat means that all derivatives at 0 are 0.)

Therefore, we can answer: the prime ideal is the set of all germs of smooth functions that are flat at 0.

But note: the problem asks for an explicit prime ideal. So we can write:

Let P = { f ∈ A | f^{(n)}(0) = 0 for all n ∈ N }.

Then P is a prime ideal.

However, we must ensure that it is not one of the excluded ones: it is not 0 because there are non-zero flat functions (like exp(-1/x^2)). It is not the whole ring because there are non-flat functions (like the identity function x). It is not the ideal {f: f(0)=0} because x is in that ideal but not in P (since its first derivative is 1, not 0).

So this should be the answer.