Classify all Maximal Ideals on this Ring

[Oster]

Classify all maximal ideals on the commutative ring of continuous functions on [0,1] (C[0,1],+,.)

I am not too confident about my solution. Can someone please look through it?

Fortunate guess:-
The set J(y) = {f in C[0,1] / f(y)=0} where y is fixed in [0,1]

This is an ideal because
(1) If g,h belong to J(y) then g-h belongs to J(y) => (J(y),+) is a subgroup of (C[0,1],+)
(2) If g is in J(y) and h is in C[0,1] then g.h is in J(y)

Also, C[0,1]/J(y) is a field because,
If J(y)+f is in C[0,1]/J(y) and J(y)+f =/= J(y), then f(y) =/= 0.
Let g(x)=1/f(y) for all x in C[0,1]
Then, (J(y)+g).(J(y)+f)=(J(y)+f.g)=(J(y)+1) [because f.g-1 is in J(y)]
Hence (J(y)+f) is invertible.

And, C[0,1]/J(y) is a field => J(y) is a maximal ideal.

To show there are no other maximal ideals:-

Let I be an ideal of C[0,1] such that I is not {0} and I is not a subset of any J(y)

That implies, for each z in [0,1], there exists f in I such that f(z) =/= 0.
multiplying this f by an appropriate constant function AND an appropriate continuous piecewise linear function will guarantee, for each z, the existence of a function g in I such that g(z)>1 AND g>0 on [0,1].

for z in [0,1], the corresponding function g and e=1/2, there exists a number d(z)>0 such that in B(z,d(z)) g is always greater than 1/2.

B(z,d(z)) is an open cover for [0,1] so it must admit a finite subcover. This will give me a finite number of functions g1, g2,..., gk in I such that their sum is a function in I that never becomes zero. Multiplying this function by its reciprocal will imply that the identically 1 function is in I. Which implies I=C[0,1]

Hence the J(y)'s are the only maximal ideals in C[0,1] (?)

Thank you for reading the whole thing.

 

[mighty2000]

Your solution is almost correct, but there are a few minor errors. Let me clarify and correct them for you.

First of all, your definition of J(y) is not quite accurate. It should be J(y) = {f in C[0,1] / f(y) = 0} where y is fixed in [0,1]. This means that J(y) is the set of all continuous functions on [0,1] that vanish at the point y. This is indeed an ideal of C[0,1], which you have correctly shown using the two properties.

Next, you have correctly shown that C[0,1]/J(y) is a field, but you have not shown that it is isomorphic to the field of real numbers. This is an important step, as it shows that J(y) is a maximal ideal. You can do this by defining a map from C[0,1]/J(y) to the real numbers, where the image of f in C[0,1]/J(y) is f(y). This map is well-defined and is an isomorphism, which proves that C[0,1]/J(y) is isomorphic to the field of real numbers.

To show that there are no other maximal ideals, your idea is correct, but the execution is not entirely accurate. Here is a simpler way to prove this:

Assume that there exists an ideal I of C[0,1] such that I is not {0} and I is not a subset of any J(y). This means that there exists a function g in I such that g is not in any J(y). This implies that g is non-zero at every point in [0,1].

Now consider the set J(g) = {f in C[0,1] / fg = 0}. This is an ideal of C[0,1], since if f and h are in J(g), then fgh = 0 and f+h is in J(g). Also, g is not in J(g) since g is non-zero at every point in [0,1]. This means that J(g) is a proper ideal of C[0,1] that strictly contains I.

However, we have already shown that the J(y)'s are the only maximal ideals in C[0,1]. This means that J(g) cannot be a maximal ideal, which contradicts our assumption that I is