We are considering the ring A of germs at 0 of smooth functions of one variable.
We are to find a prime ideal other than 0, A, and the ideal of functions vanishing at 0 (which is denoted by I, where I = {f in A | f(0)=0}).

Recall that A is the set 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 know that the ideal I = {f in A | f(0)=0} is a prime ideal because A/I is the ring of germs of smooth functions at 0 that are constant at 0, which is isomorphic to R (the real numbers). Since R is an integral domain, I is prime.

We are to find another prime ideal. One common way is to consider the ideal of functions that vanish to higher order at 0. However, note that the ideal of functions vanishing to order at least 1 is I, which we already know. The next ideal is the ideal of functions vanishing to order at least 2, but note that this ideal is not prime because it is contained in I and the quotient ring A/I is R, which is a field, so the only prime ideals are (0) and I in that quotient. Thus we need a different approach.

Another idea: consider the ideal generated by a single irreducible element. However, note that the ring A is not a UFD in the usual sense because it is not a domain? Actually, A is an integral domain because the product of two non-zero functions that are non-zero in a neighborhood of 0 is non-zero. But we are looking for a prime ideal that is not the whole ring and not the zero ideal and not I.

Alternatively, we can consider the ideal of functions that vanish at 0 and also satisfy some additional condition. However, we must ensure that the ideal is prime.

Let me recall that in the ring of germs of smooth functions, the prime ideals are in one-to-one correspondence with the prime ideals in the ring of formal power series R[[x]]. This is because by the Hadamard preparation theorem, every germ of a smooth function can be written as a unit times a formal power series (in the smooth case, we have the formal Taylor series). However, note that the ring of formal power series R[[x]] is a discrete valuation ring (DVR) with the maximal ideal (x). The prime ideals of R[[x]] are (0) and (x^n) for n>=1? Actually, no: the prime ideals of R[[x]] are (0) and the ideal generated by x, because R[[x]] is an integral domain and (x) is prime. But note that (x) is the only non-zero prime ideal? Actually, that's not true: for example, the ideal (x^2) is not prime because it is not radical? Actually, in a DVR, the non-zero prime ideals are the powers of the maximal ideal? No, in a DVR, the non-zero prime ideals are all of the form (π^k) for k>=1, but note that in R[[x]], the maximal ideal is (x) and the ideals are (x^n) for n>=0. However, (x^2) is not prime because (x^2) contains x^2 and x*x = x^2 is in (x^2), but x is not in (x^2). So (x^2) is not prime. The only prime ideals in R[[x]] are (0) and (x).
But wait, we are in the smooth germs ring. The ring A is actually a local ring with maximal ideal I (the ideal of functions vanishing at 0). Moreover, A is a Noetherian ring? Actually, it is not Noetherian. However, we know that the prime ideals of A correspond to the prime ideals of the associated graded ring. But perhaps we can use the fact that the ideals of A are in bijection with the ideals of the formal power series ring via the Taylor series.

However, note that the ring A is a DVR? Actually, it is not a DVR because the maximal ideal I is not principal. For example, there is no single function that generates I because if we take f(x)=x and g(x)=x^2, then g is in I but not in the ideal generated by f. So A is not a DVR.

Another approach: consider the ideal of functions that vanish at 0 and have derivative 0 at 0. That is, the ideal J = {f in A | f(0)=0 and f'(0)=0}.

Let's check if J is prime. Suppose we have two functions f and g such that fg is in J. Then (fg)(0)=0 and (f*g)'(0)=0. We want to show that either f is in J or g is in J.

But note: (f*g)(0)=f(0)g(0)=0, so either f(0)=0 or g(0)=0. Without loss of generality, assume f(0)=0. Now, the derivative: (fg)'(0) = f'(0)*g(0) + f(0)*g'(0) = f'(0)*g(0) = f'(0)g(0). But we have (fg)'(0)=0, so f'(0)*g(0)=0. Since f(0)=0 and we are assuming g(0) might not be 0? Actually, if g(0) is not zero, then we must have f'(0)=0. But then f would be in J? Not exactly: we have f(0)=0 and f'(0)=0, so f is in J. But what if g(0)=0? Then we don't know about g'(0). Actually, if g(0)=0, then g might not be in J because we don't know about g'(0). So we must also show that if g(0)=0 then we can deduce that g is in J? But we don't have that.

Counterexample: Consider f(x)=x and g(x)=x. Then fg=x^2. Now, x^2 is in J because (x^2)(0)=0 and (x^2)'(0)=0. But neither f nor g is in J: because f(0)=0 but f'(0)=1≠0. Similarly for g. So this is a counterexample to the prime condition? Actually, we have fg in J but neither f nor g is in J. Therefore, J is not prime.

So that ideal doesn't work.

Another idea: consider the ideal of functions that vanish at 0 and have all derivatives at 0 equal to 0. But note that the ring of smooth functions has the property that the only function with all derivatives 0 at a point is the zero function? Actually, no: the function f(x)=e^{-1/x^2} for x≠0 and f(0)=0 is smooth and all derivatives at 0 are 0. So we can consider the ideal of functions that are flat at 0, i.e., all derivatives at 0 are 0.

Let J be the set of functions in A that are flat at 0 (i.e., f and all its derivatives vanish at 0). Then J is an ideal. Is it prime?

Suppose fg is in J. Then (fg)(0)=0 and (fg)'(0)=0, ... and all higher derivatives of fg vanish at 0. We want to show that either f is in J or g is in J.

However, we can have counterexamples: consider f(x)=e^{-1/x^2} and g(x)=x. Then f is flat at 0 (so in J) and g is not. Then fg is flat? Let h(x)=f(x)g(x)=x * e^{-1/x^2}. Then h(0)=0, and 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 derivative h'(0) is 0? Actually, we must compute the derivative at 0 by the definition:
h'(0) = lim_{x->0} (h(x)-h(0))/x = lim_{x->0} e^{-1/x^2} * (1/x)
which is 0 because the exponential decays faster than any polynomial. Similarly, all higher derivatives of h will be 0 at 0. So h is flat. But note that g(x)=x is not flat. So in this case, we have f in J and g not in J, and fg in J. This does not contradict primality. But we need to see if we can have both f and g not in J but fg in J.

Consider f(x)=e^{-1/x^2} and g(x)=e^{-1/x^2}. Then both are flat, so they are in J. Now consider f(x)=e^{-1/x^2} and g(x)=x^2. Then g is not flat? Actually, g(x)=x^2: g(0)=0, g'(0)=0, but g''(0)=2≠0. So g is not flat. Then fg = x^2 * e^{-1/x^2}. Is this flat? We can compute the derivatives: they will be a combination of polynomials in 1/x times e^{-1/x^2}. The limit as x->0 of any such term is 0. So fg is flat. But note that g is not flat. So again, we have one in J and one not.

Now, can we have two non-flat functions whose product is flat? Let f(x)=x and g(x)=x. Then f*g=x^2. Is x^2 flat? We know that the second derivative of x^2 is 2, which is not 0. So x^2 is not flat. So that doesn't work.

What about f(x)=x and g(x)=e^{-1/x^2}? Then f*g = x * e^{-1/x^2}. We have already seen that this is flat. But note that f(x)=x is not flat (since its first derivative is 1 at 0) and g is flat. So again, one is flat and the other is not.

How about two non-flat functions? Let f(x)=x and g(x)=x^2. Then f*g=x^3. This is not flat because the third derivative is 6≠0.

Let f(x)=x and g(x)=e^{-1/x^2} + x. Then g is not flat because g(0)=0 but g'(0) = limit_{x->0} (g(x)-g(0))/x = (e^{-1/x^2}+x)/x = e^{-1/x^2}/x + 1. The term e^{-1/x^2}/x goes to 0, so g'(0)=1. Then fg = x(e^{-1/x^2}+x) = xe^{-1/x^2} + x^2. This is flat? The function xe^{-1/x^2} is flat and x^2 is not. But the sum of a flat function and a non-flat function is not flat? Actually, we need to check the derivatives of fg at 0. The function x^2 has derivatives that are non-zero at 0 for order 2 and higher. However, the function xe^{-1/x^2} has all derivatives 0. So the sum will have the same derivatives as x^2 at 0? That is, the derivatives of x^2 are non-zero for order at least 2. So f*g is not flat.

After several attempts, I recall that in the ring of germs of smooth functions, the prime ideals are the following:

• The zero ideal.
• The ideal I of functions vanishing at 0.
• The ideal J of functions that are flat at 0 (i.e., all derivatives vanish at 0).

And also, there are prime ideals that are not principal. In fact, the ideal J is prime. Why?

Suppose that fg is in J. Then (fg) is flat at 0. We want to show that either f is flat or g is flat. We can use the following fact: if a function is flat at 0, then its Taylor series at 0 is 0. Conversely, if the Taylor series is non-zero, then the function is not flat. But note that the ring A has the property that the product of two non-flat functions can be flat? We have not found a counterexample and in fact, it is true that J is prime.

Proof: Suppose that f and g are not flat. Then there exists some integer n such that f^{(n)}(0) ≠ 0 and some integer m such that g^{(m)}(0) ≠ 0. Let k = max(n, m). Then the Taylor polynomial of f at 0 of degree k is non-zero and the Taylor polynomial of g at 0 of degree k is non-zero. Since the Taylor polynomial of fg at 0 is the convolution of the Taylor polynomials of f and g, which is non-zero. Therefore, fg is not flat. This contradiction shows that if f*g is flat, then at least one of f or g must be flat.

Therefore, J is a prime ideal.

But note: we must also check that J is not equal to I. Indeed, I contains functions that are not flat, such as x. And J is a proper subset of I because it does not contain x.

Also, J is not the zero ideal because it contains the function e^{-1/x^2} (which is non-zero).

Therefore, we have found a prime ideal: J, the ideal of flat functions at 0.

But note: the problem asks for a prime ideal other than 0, A, and I. We have found J.

However, we must make sure that J is not equal to I. We have already argued that it is a proper subset. Also, it is non-zero.

So the answer is J = {f in A | f(0)=0 and f^{(k)}(0)=0 for all k>=1}.

But note: the function f(x)=e^{-1/x^2} for x≠0 and f(0)=0 is in J. Also, any function that is identically zero in a neighborhood of 0 is in J. But note that there are other functions: for example, the function f(x)=x^2 is not in J because f''(0)=2≠0.

Therefore, we have found the prime ideal J.

However, note that the problem says "germs at 0 of smooth functions". The definition of J in terms of the derivatives at 0 is well-defined for germs: because if two functions are equivalent (i.e., agree on some neighborhood of 0), then they have the same values and the same derivatives at 0.

So we can define J as the set of germs f such that f(0)=0 and for every positive integer k, the k-th derivative of the function at 0 is 0.

Hence, we have the prime ideal J.

But wait: is J really an ideal? We need to check that it is closed under addition and under multiplication by any element of A. Closure under addition: if f and g are flat, then f+g is flat? Yes, because the derivatives of f and g are 0 at 0, so the derivatives of f+g are 0. Closure under multiplication by an arbitrary element h in A: we need to show that if f is flat and h is any germ, then hf is flat. This is true because the product of a flat function and any smooth function is flat. Why? The Taylor series of hf is the Taylor series of h times the Taylor series of f. But the Taylor series of f is 0, so the product is 0. Therefore, h*f is flat.

Therefore, J is an ideal.

In conclusion, we have found the prime ideal J.

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

J = { f ∈ A | f(0) = 0 and for all k ≥ 1, f^{(k)}(0) = 0 }

However, we must be cautious: the function that is identically zero in a neighborhood of 0 is in J, and also the function e^{-1/x^2} is in J. But note that the function e^{-1/x^2} is not zero at any point, but it is zero at 0 and all derivatives are 0.

Therefore, we have found the prime ideal J.

But note: there is a subtlety. The ring A is the ring of germs. The ideal J is the set of germs of functions that are flat at 0. This is a well-defined ideal.

So the answer is the ideal J as defined.

However, note that the problem says "other than 0, A, or {f ∈ A | f(0)=0}". So we have excluded the zero ideal, the whole ring, and the ideal I. The ideal J is none of these.

Therefore, we have the solution.

首先，用户的问题是：为什么函数 $f(x)=e^{-1/x^2}$ 对于 $x\ne 0$ 且 $f(0)=0$ 是光滑函数。光滑函数通常意味着无穷次可微，即所有阶导数都存在。

我需要证明这个函数在 $x=0$ 处是无穷次可微的。用户给出的定义是：当 $x\ne 0$ 时，$f(x)=e^{-1/x^2}$，而当 $x=0$ 时，$f(0)=0$。

为了证明它是光滑的，我必须检查所有阶导数在 $x=0$ 处是否存在且连续。实际上，对于光滑性，我们需要所有导数存在且连续，但在这个例子中，由于函数在 $x\ne 0$ 时是无限可微的，关键是检查在 $x=0$ 处的行为。

让我先考虑低阶导数。设 $t=1/x$，那么当 $x\to 0$ 时，$t\to \infty$。函数变为 $f(x)=e^{-t^2}$。

现在，考虑在 $x=0$ 处的导数。一阶导数：$f^{\prime }(x)$ 对于 $x\ne 0$。

计算一阶导数：$f^{\prime }(x)=e^{-1/x^2}\cdot \frac{d}{dx}(-1/x^2)=e^{-1/x^2}\cdot (2/x^3)$。

类似地，二阶导数：等等。

现在，在 $x=0$ 处，定义导数：$f^{\prime }(0)={\lim }_{h\to 0}\frac{f(h)-f(0)}{h}={\lim }_{h\to 0}\frac{e^{-1/h^2}}{h}$。

令 $h=1/t$，当 $h\to 0$，$t\to \infty$，所以 $\frac{e^{-t^2}}{1/t}=t{e}^{-t^2}$。

现在，${\lim }_{t\to \infty }t{e}^{-t^2}$。由于指数衰减比多项式增长快，这个极限是 0。

类似地，对于更高阶导数，它们都会包含 $x$ 的正幂次乘以 $e^{-1/x^2}$，当 $x\to 0$ 时，指数项主导，极限为 0。

更精确地，对于任何 $n$，第 n 阶导数在 $x=0$ 处的极限是 0。

证明所有导数在 0 处存在且为 0。

首先，对于 $x\ne 0$，$f(x)=e^{-1/x^2}$，其导数可以通过链式法则计算。

设 $g(u)=e^{-u}$，其中 $u=1/x^2$，那么 $g^{\prime }(u)=-e^{-u}$，$u^{\prime }=-2/x^3$，所以 $f^{\prime }(x)=g^{\prime }(u)u^{\prime }=(-e^{-u})\cdot (-2/x^3)=(2/x^3)e^{-1/x^2}$。

类似地，二阶导数：$f^{\prime \prime }(x)=\frac{d}{dx}\left(\frac{2}{x^3}{e}^{-1/x^2}\right)$。

使用乘积法则：$\frac{d}{dx}(2x^{-3})e^{-1/x^2}+2x^{-3}\frac{d}{dx}(e^{-1/x^2})$。

$\frac{d}{dx}(2x^{-3})=-6x^{-4}$，而 $\frac{d}{dx}(e^{-1/x^2})=(2/x^3)e^{-1/x^2}$ 如上所述。

所以 $f^{\prime \prime }(x)=-6x^{-4}{e}^{-1/x^2}+2x^{-3}\cdot (2/x^3)e^{-1/x^2}=-6x^{-4}{e}^{-1/x^2}+4x^{-6}{e}^{-1/x^2}$。

类似地，更高阶导数将包含 $x$ 的正幂次和负幂次乘以 $e^{-1/x^2}$。

关键点是，对于任何整数 $m\ge 0$，在 $x=0$ 处的第 m 阶导数定义为：

f^{(m)}(0) = \lim_{x \to 0} \frac{f^{(m)}(x)} ，但由于在 0 处可能未定义，我们需要使用极限。

实际上，定义是：如果极限存在，则 $f^{(m)}(0)={\lim }_{h\to 0}\frac{f^{(m-1)}(h)-f^{(m-1)}(0)}{h}$，但我们需要递归地做。

由于所有导数在 0 处似乎都是 0，我们可以证明对于每个 $m$，$f^{(m)}(0)=0$。

考虑函数的形式。注意，对于 $x\ne 0$，$f^{(m)}(x)$ 可以写成 $p_m(1/x)e^{-1/x^2}$，其中 $p_m$ 是某个多项式。

例如，对于 m=0，$f(x)=e^{-1/x^2}$，即 $p_0(t)e^{-t^2}$ 其中 $t=1/x$，$p_0(t)=1$。

对于 m=1，$f^{\prime }(x)=(2/x^3)e^{-1/x^2}=2t^3{e}^{-t^2}$，因为 $t=1/x$，所以 $x=1/t$，$1/x^3=t^3$，所以 $f^{\prime }(x)=2t^3{e}^{-t^2}$。

类似地，对于 m=2，之前有 $f^{\prime \prime }(x)=(-6x^{-4}+4x^{-6})e^{-1/x^2}=(-6t^4+4t^6)e^{-t^2}$。

一般来说，$f^{(m)}(x)=q_m(t)e^{-t^2}$，其中 $q_m(t)$ 是多项式，且当 $t\to \infty$ 时，由于指数衰减，$q_m(t)e^{-t^2}\to 0$。

更精确地，对于任何多项式 $p(t)$，有 ${\lim }_{t\to \infty }p(t)e^{-t^2}=0$。

因此，对于每个固定的 $m$，当 $x\to 0$（即 $t\to \infty$）时，$f^{(m)}(x)\to 0$。

现在，在 $x=0$ 处，导数定义为极限。例如，一阶导数：$f^{\prime }(0)={\lim }_{h\to 0}\frac{f(h)-f(0)}{h}={\lim }_{h\to 0}\frac{e^{-1/h^2}}{h}={\lim }_{t\to \infty }t{e}^{-t^2}=0$。

类似地，对于二阶导数：$f^{\prime \prime }(0)={\lim }_{h\to 0}\frac{f^{\prime }(h)-f^{\prime }(0)}{h}$。但 $f^{\prime }(0)$ 是多少？我们刚刚证明它是 0。而 $f^{\prime }(h)$ 对于 $h\ne 0$ 是 $(2/h^3)e^{-1/h^2}$，所以

$f^{\prime \prime }(0)={\lim }_{h\to 0}\frac{(2/h^3)e^{-1/h^2}-0}{h}={\lim }_{h\to 0}\frac{2}{h^4}{e}^{-1/h^2}={\lim }_{t\to \infty }2t^4{e}^{-t^2}$。

由于 $e^{-t^2}$ 比任何多项式衰减得更快，所以这个极限是 0。

类似地，对于任何 $m$，$f^{(m)}(0)={\lim }_{h\to 0}\frac{f^{(m-1)}(h)-f^{(m-1)}(0)}{h}$。

由归纳法，假设 $f^{(k)}(0)=0$ 对于 $k<m$ 成立。那么

$f^{(m)}(0)={\lim }_{h\to 0}\frac{f^{(m-1)}(h)}{h}$，因为 $f^{(m-1)}(0)=0$。

现在，$f^{(m-1)}(h)$ 对于 $h\ne 0$ 是 $p_{m-1}(1/h)e^{-1/h^2}$，其中 $p_{m-1}$ 是多项式。设 $p_{m-1}(t)=a_k{t}^k+\cdots$，其中 $k$ 是某个度数。

实际上，从之前的计算中，对于每个导数，多项式次数增加。例如，$f^{(m)}(x)$ 包含 $x$ 的负幂次，因此当用 $t=1/x$ 表示时，它是 $t$ 的正幂次。

具体来说，$f^{(m)}(x)=q_m(1/x)e^{-1/x^2}$，其中 $q_m$ 是多项式，且当 $t\to \infty$ 时，$q_m(t)e^{-t^2}\to 0$。

现在，$f^{(m)}(0)={\lim }_{h\to 0}\frac{f^{(m-1)}(h)}{h}={\lim }_{h\to 0}\frac{q_{m-1}(1/h)e^{-1/h^2}}{h}={\lim }_{t\to \infty }t\cdot q_{m-1}(t)e^{-t^2}$。

因为 $h=1/t$，所以 $1/h=t$，因此 $\frac{f^{(m-1)}(h)}{h}=t\cdot q_{m-1}(t)e^{-t^2}$。

而 $q_{m-1}(t)$ 是多项式，所以 $t{q}_{m-1}(t)$ 也是多项式，设为 $r_m(t)$，那么 $r_m(t)e^{-t^2}\to 0$ 当 $t\to \infty$。

因此，对于所有 $m$，极限是 0，所以 $f^{(m)}(0)=0$。

此外，由于对于 $x\ne 0$，导数存在且连续，并且当 $x\to 0$ 时，$f^{(m)}(x)\to 0=f^{(m)}(0)$，所以导数在 0 处连续。

因此，所有阶导数都存在且连续，所以函数是无穷次可微的，即光滑的。

在上下文中，这个函数被用作在 $x=0$ 处为平顺的但在更高导数中“消失”的函数示例，正如在素理想问题中所示。

所以，总结：$f(x)=\{\begin{array}{ll}{e}^{-1/x^2}& x\ne 0\\ 0& x=0\end{array}$ 是光滑的，因为所有在 0 处的导数都存在且等于 0，并且与邻域内的导数相匹配。