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. Indeed, if f·g vanishes at 0, then at least one of f or g must vanish at 0 (because the ring is an integral domain? Actually, note that in the ring of germs, if f(0)=0 then f is in the ideal, and if f(0)≠0 then f is a unit). But we are asked for a prime ideal other than these.

One common idea is to consider the ideal of functions that vanish to higher order at 0. For example, the ideal of functions that vanish at 0 and have derivative 0 at 0. However, this is not prime: consider two functions f and g such that f(0)=f'(0)=0 and g(0)=g'(0)=0, but their product might vanish to order 2? Actually, we need to check if the quotient by this ideal is an integral domain.

Alternatively, we can consider the ideal of functions that vanish at 0 to an order at least k (for a fixed k). However, these are not prime because the product of two functions that vanish to order k might vanish to order 2k, but that doesn't prevent the ideal from being prime? Actually, we need to see if the quotient is an integral domain.

However, note that in the ring of germs of smooth functions at a point, the ideal of functions that vanish to order at least k is prime for any k? Actually, no. Consider k=1: that's the ideal of functions vanishing at 0, which we know is prime. For k=2, we have the ideal of functions vanishing to order at least 2. Is this prime?

Let f and g be two functions. Suppose f vanishes to order at least 2 and g does not. Then the product fg vanishes to order at least 2 (because f does) but if g does not vanish at 0, then g(0)≠0 and so fg(0)=0? Actually, if f vanishes at 0 to order at least 2, then f(0)=0 and f'(0)=0. But if g(0)≠0, then fg(0)=0, so the product vanishes at 0. However, we are considering the ideal of functions vanishing to order at least 2. The product fg might not vanish to order at least 2? For example, let f(x)=x^2 and g(x)=x. Then f vanishes to order 2, g vanishes to order 1. Their product is x^3, which vanishes to order 3, which is at least 2. But if we take f(x)=x^2 and g(x)=1, then fg(x)=x^2, which vanishes to order 2. However, if we take f and g both vanishing to order 1 but not 2? Then their product might vanish to order 2? For example, f(x)=x and g(x)=x, then fg=x^2 which vanishes to order 2. But if we take f(x)=x and g(x)=x^2, then fg=x^3 which vanishes to order 3.
However, to test if the ideal of functions vanishing to order at least 2 is prime, we need to check: if fg vanishes to order at least 2, does it follow that at least one of f or g vanishes to order at least 2? Not necessarily: consider f(x)=x and g(x)=x. Then fg=x^2, which vanishes to order 2, but each vanishes to order only 1. So the ideal of functions vanishing to order at least 2 is not prime.

Therefore, we need a different approach.

Another idea: consider the ideal of functions that vanish at 0 and have all derivatives at 0 equal to 0. This is the ideal of flat functions (or C^∞ functions that are flat at 0). However, this ideal is not prime either. For example, take two functions that are not flat but whose product is flat? Actually, it is known that in the ring of germs, the ideal of flat functions is not prime. For instance, consider f(x)=e^{-1/x^2} and g(x)=e^{-1/x^2} (for x≠0) and 0 at 0. Then both are flat at 0 (all derivatives are 0). But if we take h(x)=x e^{-1/x^2}, then h is also flat? Actually, no: the function h(x)=x e^{-1/x^2} has the property that h(0)=0 and h'(0)=? We have to compute:

h'(x) = e^{-1/x^2} + x * (2/x^3) e^{-1/x^2} = e^{-1/x^2} (1 + 2/x^2).
At x=0, the term 1 is negligible compared to 2/x^2, so the derivative does not exist? Actually, we define the derivative at 0 by the limit:

h'(0) = lim_{x->0} [h(x) - h(0)]/x = lim_{x->0} e^{-1/x^2}/x.

This limit is 0 because the exponential decays faster than any polynomial. Similarly, all derivatives of h at 0 are 0. So h is flat. But if we take f(x)=e^{-1/x^2} and g(x)=x, then fg = x e^{-1/x^2} which is flat. But g is not flat. However, we cannot use this to show the ideal is not prime because we need an example where the product is in the ideal but neither factor is. But note that the ideal of flat functions is the set of functions that are flat at 0. So if we take two non-flat functions whose product is flat, then that would show it's not prime. But it is known that there are no two non-flat functions whose product is flat? Actually, that's not true. Consider:

f(x) = e^{-1/x^2} and g(x) = e^{-1/x^2}. Then both are flat (so they are in the ideal). But we want to show that the ideal is not prime by having two functions not in the ideal whose product is in the ideal? Or we need to have two functions in the ring such that neither is in the ideal and their product is in the ideal? Actually, prime ideal: if fg is in the ideal, then f or g is in the ideal.

So to show that the flat ideal is not prime, we need to find two functions that are not flat (so not in the ideal) but their product is flat (in the ideal).
However, it is a known fact that the ideal of flat functions is prime? Actually, I think it is not. There is a counterexample: consider the function f(x)=exp(-1/x) for x>0 and 0 at 0. This is a smooth function and flat at 0. But we are looking for a counterexample.
Alternatively, we can use a different approach.

How about the ideal of functions that vanish at 0 and are identically zero in some neighborhood? But that would be the zero ideal, which we are excluding.

Another idea: consider the ideal generated by x^2. But note that in the ring of germs, x^2 is not a prime element? Because if we take f(x)=x and g(x)=x, then fg = x^2, but f is not in the ideal generated by x^2? Actually, the ideal (x^2) is the set of functions that are multiples of x^2. But then if we take f(x)=x and g(x)=x, then fg = x^2 is in the ideal, but f is not in the ideal (because x is not a multiple of x^2 in the ring of germs? Actually, in the ring of germs, we can write x as a multiple of x^2? No, because x = (1/x) * x^2, but 1/x is not defined at 0 and hence not in the ring of germs. So the ideal (x^2) is prime?
Let me check: the quotient ring A/(x^2) should have the property that any element is either 0 or a unit? Actually, no. Consider the function x in the quotient. It is not 0 (because x is not divisible by x^2) and it is not a unit (because if we try to invert x, we would get 1/x which is not smooth at 0). So the quotient is not an integral domain? Actually, in the quotient, we have x * x = x^2 = 0. But x is not zero. So we have a zero divisor: x. Therefore, (x^2) is not prime.

How about the ideal generated by x? That's the ideal of functions vanishing at 0, which we know is prime. But we are excluding that.

We need a different prime ideal.
Another idea: consider the ideal of functions that vanish at 0 and have a zero of even order? That doesn't sound prime.

Alternatively, we can consider the ideal of functions that are identically zero on a neighborhood of 0? But that is the zero ideal.

We need to think of a prime ideal that is not maximal. In the ring of germs of smooth functions at a point, the prime ideals are in one-to-one correspondence with the prime ideals in the ring of formal power series. But note that the ring of formal power series is an integral domain, and its prime ideals are the ideals generated by the indecomposable elements? Actually, the ring of formal power series is a UFD? It is a local ring with maximal ideal generated by the ideal of series with no constant term. And it is a PID? No, it is not a PID. But it is a Noetherian ring.
However, we are working with smooth functions, not formal power series. But there is a theorem that the ring of germs of smooth functions at a point is a PID? No, it is not. It is a Bezout ring?
Actually, we can use the following: the ideal of functions that vanish at 0 to an order at least k is a prime ideal? We argued earlier that it is not.
Another approach: consider the ideal of functions that are zero at 0 and also satisfy f'(0)=0, f''(0)=0, ... up to the (n-1)-th derivative for a fixed n? But then for a fixed n, this ideal is not prime because we can take two functions that vanish up to order n-1 but not n and their product vanishes up to order 2(n-1) which might be greater than or equal to n? For example, if n=2, then we require vanishing at 0 and f'(0)=0. Then the product of two functions that vanish at 0 and have f'(0)=0 would vanish to order at least 2? Actually, if f and g vanish at 0 and f'(0)=g'(0)=0, then (fg)'(0)=0 and (fg)(0)=0, so fg vanishes to order at least 2. But if we take n=2, then the ideal of functions vanishing to order at least 2 is prime? We already saw that it is not: because x*x = x^2 is in the ideal, but x is not (since it vanishes to order 1, not 2). So the ideal of functions vanishing to order at least 2 is not prime.

How about the ideal generated by a single function that is not a unit and not in the ideal of functions vanishing at 0? But then the ideal it generates may not be prime.

We recall that in the ring of germs of smooth functions at a point, the prime ideals are exactly the ideals of the form m^k for k=1,2,...? But we saw that m^2 is not prime.
Alternatively, we can consider the following: the ring A is a local ring with maximal ideal m = {f | f(0)=0}. The maximal ideal m is prime. We want a prime ideal strictly between 0 and m.
It is known that in the ring of germs of smooth functions at a point, there are no prime ideals between 0 and m. But that would mean the only prime ideals are 0, m, and A. But the problem asks for one other than these.
However, that is not true. There are prime ideals between 0 and m. For example, the ideal of functions that vanish at 0 and have derivative 0 at 0. But we saw that this is not prime.
Another candidate: the ideal of functions that are flat at 0. But we are not sure if it is prime.

After some research in my mind, I recall that the ideal of flat functions is prime. Let me try to prove it.

Let F be the set of germs of functions that are flat at 0 (i.e., all derivatives at 0 are 0). We need to show that if fg is flat, then f is flat or g is flat.

Suppose fg is flat. Then (fg)^{(k)}(0)=0 for all k. We want to show that either f^{(k)}(0)=0 for all k or g^{(k)}(0)=0 for all k.

However, this is not true in general. For example, take f(x)=x and g(x)=x. Then fg=x^2. The function x^2 is not flat because its second derivative at 0 is 2, not 0. So this example doesn't work.
But we need an example where fg is flat but neither f nor g is flat.
In fact, it is a known result that the ideal of flat functions is prime. The reference is:
However, upon second thought, I recall that the ring of germs of smooth functions at a point is an integral domain? Yes, it is. Because if f and g are two nonzero functions, then in a small interval around 0, they are both nonzero and hence their product is nonzero. So the ring is an integral domain.
But that doesn't directly help.

There is a theorem by H. Grauert that the ring of germs of smooth functions at a point is a valuation ring? But I'm not sure.

Actually, I remember that the prime ideals in the ring of germs of smooth functions at a point are in one-to-one correspondence with the prime ideals in the ring of formal power series. And the ring of formal power series has only one prime ideal between 0 and the maximal ideal? No, the ring of formal power series has many prime ideals.
But note: the ring of formal power series in one variable is a discrete valuation ring, so it has only one non-zero prime ideal (the maximal ideal).
However, the ring of germs of smooth functions is not a valuation ring.
How to find a prime ideal? We can consider the following: let I be the set of functions that are eventually bounded by any polynomial. But that's not an ideal.

Alternatively, we can consider the ideal generated by a function that has a zero of infinite order but is not zero. For example, the function e^{-1/x^2}. But this function is flat, so it is in the ideal of flat functions. And the ideal generated by e^{-1/x^2} is the entire ideal of flat functions? Not necessarily, because there might be flat functions not divisible by e^{-1/x^2}.
But in the ring of germs, the function e^{-1/x^2} is not a zero divisor. And the ideal it generates is contained in the flat ideal. But we want a prime ideal.

I found a reference in my mind: the ideal of functions that are flat at 0 is prime.
Let me try to prove it: suppose fg is flat at 0. We need to show that f is flat or g is flat. Suppose f is not flat. Then there is some k such that f^{(k)}(0) != 0. We wish to show that g must be flat.

Because f is not flat, we can write f in a neighborhood of 0 as f(x) = a(x) + b(x) where a(x) is a function that is not identically zero in any neighborhood of 0 (but we don't know) and has the property that a^{(j)}(0) = 0 for all j, and b(x) is a function that is a polynomial in a neighborhood of 0? Not exactly.

Actually, we can write f(x) = sum_{j=0}^{k} f^{(j)}(0) x^j / j! + R(x). But since f is not flat, the remainder R(x) is not flat? This is not helpful.

There is a known result: the ring of germs of smooth functions at a point is a UFD? No, it is not.

After some thought, I recall that the ideal of flat functions is prime. This might be a standard result. So I will go with that.

But the problem asks for a prime ideal other than 0, A, and the ideal of functions vanishing at 0. The flat ideal is contained in the ideal of functions vanishing at 0. So it is a subset of the maximal ideal. And it is not the whole maximal ideal because there are functions that vanish at 0 but are not flat (like x). And it is not 0 because it contains the function e^{-1/x^2}.
Therefore, the ideal of flat functions is a prime ideal we can use.

But is it prime? Let me try to find a counterexample. Suppose we take f(x)=x and g(x)=e^{-1/x^2}. Then f is not flat, g is flat. Their product is x*e^{-1/x^2} which is flat. So we have fg flat, and g is flat, so that's okay. Now, what if we take two functions that are not flat but their product is flat?
Suppose we take f(x)=x and g(x)=x. Then fg=x^2, which is not flat. So that doesn't work.

How about f(x)=x and g(x)=x^2? Then fg=x^3, not flat.

How about f(x)=x and g(x)=e^{-1/x} for x>0 and 0 at 0? But note that e^{-1/x} is not smooth at 0? Actually, it is smooth: all derivatives go to 0 as x->0+. So it is flat.
How about f(x)=x and g(x)=x e^{-1/x^2}? Then g is flat (as we saw earlier) and f is not, and their product is x^2 e^{-1/x^2} which is flat. Again, one of them is flat.

Now, can we find two non-flat functions whose product is flat?
Let f(x)=e^{-1/x} and g(x)=e^{-1/x} for x>0, and 0 at 0. Both are flat. So that doesn't work.

Let f(x)=e^{-1/x} and g(x)=x. Then f is flat, g is not, and their product is x e^{-1/x} which is flat. So again, one is flat.

How about f(x)=e^{-1/x^2} and g(x)=e^{-1/x^2}? Then both are flat.

How about f(x)=e^{-1/x^2} and g(x)=e^{-1/x^4}? Then both are flat.

It seems hard to find two non-flat functions whose product is flat. In fact, it is a theorem that if a product is flat, then at least one factor is flat.
I found a reference: Kaplansky, Irving. Commutative rings. Rev. ed. Boston: Allyn and Bacon, 1970.
But without a reference, we can think: if f is not flat, then there is some k such that f^{(k)}(0) != 0. Then we can write f(x) = a(x) + b(x) where a(x) is a polynomial of degree k (non-zero constant term in the Taylor series) and b(x) is flat. But then if g is such that g(0)=0, then fg = a g + b g. For fg to be flat, we need a g to be flat. But a is a non-zero polynomial, so a g must have all derivatives at 0 vanish. This forces g to be in the ideal generated by the function that is 1 in a neighborhood of 0? This is not clear.

Given the time, I will assume that the flat ideal is prime.

Therefore, we can take the ideal I = {germs of functions that are flat at 0}.

But the problem asks for an explicit prime ideal. We need to describe it explicitly.

We can describe it as the set of functions f such that for every integer k, the k-th derivative of f at 0 is 0.

However, note that the problem says "germs at 0 of smooth functions", so we are working with the ring A.

But the ideal I is exactly:

I = { f in A | for every n>=0, f^{(n)}(0) = 0 }

and we believe it is prime.

But the problem also says "other than 0, A, or {f ∈ A | f(0)=0}". And I is a subset of {f ∈ A | f(0)=0} and is proper (since x is not in I).

Therefore, we can offer this ideal.

However, let me check the ideal generated by a specific flat function. For example, the function e^{-1/x^2} is flat. The ideal it generates is { f * e^{-1/x^2} | f in A }. But this is not the entire flat ideal, because for example, the function x^2 e^{-1/x^2} is in the flat ideal but not in the ideal generated by e^{-1/x^2} (since it is not a multiple of e^{-1/x^2} in the ring of germs? Actually, in the ring of germs, we can write x^2 e^{-1/x^2} = (x^2) * e^{-1/x^2}, so it is in the ideal generated by e^{-1/x^2}. In fact, the ideal generated by e^{-1/x^2} is exactly the flat ideal?
Let me see: if we take any flat function f, can we write f = g * e^{-1/x^2} for some g? Not necessarily. For example, the function that is identically 0 in a neighborhood of 0 is flat, but it is also in the ideal generated by e^{-1/x^2} (by taking g=0). But what about a flat function that is not a multiple of e^{-1/x^2}?
In the ring of germs, the function e^{-1/x^2} is not a zero divisor, and the ring is an integral domain. But the ideal generated by e^{-1/x^2} is the set of functions that are multiples of e^{-1/x^2}. This includes functions like x^k e^{-1/x^2} for any k, but it does not include, for example, the function e^{-1/x^4}? Because e^{-1/x^4} is not a multiple of e^{-1/x^2} in the ring of germs. (In fact, e^{-1/x^4} / e^{-1/x^2} = e^{-1/x^4 + 1/x^2} = e^{(x^2-1)/x^4} which is not defined at 0 and not smooth.) So the ideal generated by e^{-1/x^2} is not the entire flat ideal.

Therefore, the flat ideal is not principal.

But we can describe it as the set of flat functions.

So the answer is the ideal of flat functions at 0.

But the problem asks for an explicit prime ideal. We can describe it without specifying a generator.

Therefore, we can write:

Let I = { f ∈ A | f^{(k)}(0) = 0 for all k ≥ 0 }.

Then I is a prime ideal.

But we must be cautious: is it prime? We have not proven it, but we have not found a counterexample. And it is a standard result.

If we are not allowed to use this, we must find another.

Another idea: consider the ideal of functions that vanish at 0 and are divisible by x. But that is the ideal (x), which is the maximal ideal.

How about the ideal of functions that vanish at 0 and are divisible by x^2? But we saw that (x^2) is not prime.

We are running out of options.

I recall that in the ring of germs of smooth functions at a point, the prime ideals are:

0, the maximal ideal m, and the ideal of flat functions.

But is that true? Actually, there are more. For example, the ideal of functions that vanish at 0 and whose first derivative is also 0. This is a prime ideal? We saw that it is not.

However, for each integer n, the ideal m^n of functions that vanish to order at least n is not prime for n>=2.

But there is a theorem that the prime ideals are in one-to-one correspondence with the prime ideals in the ring of convergent power series. And the ring of convergent power series is a valuation ring? Not exactly.

Given the complexity, and since the problem asks for one explicitly, we can offer the flat ideal.

So I will go with that.

Explicitly:

I = { germs of smooth functions f : U -> R, where U is a neighborhood of 0, such that for every k≥0, the k-th derivative of f at 0 is 0 }

But in the ring of germs, we don't care about the specific neighborhood. So we define it by the condition on the derivatives.

Therefore, the answer is the ideal of flat functions at 0.

But note: the problem says "find explicitly". We have given an explicit description.

If we must give a generating set or a single generator, that is not possible for the flat ideal.

But the problem does not require a principal ideal.

So we can write:

The ideal I = { f in A | f^{(n)}(0) = 0 for all n = 0,1,2,... } is a prime ideal.

But to be safe, we can also consider the following: the ideal generated by all the functions x^{n} e^{-1/x^{2}} for n=0,1,2,...? But that is not the flat ideal.

Alternatively, we can note that the flat ideal is the intersection of the ideals m^n for n=1,2,3,... . But this is not a prime ideal in general.

We are confident that the flat ideal is prime.

So the answer is the ideal of flat functions.