1. Let $X$ be a Hausdorff space and $C(X)$ the ring of complex-valued continuous functions. Let $M_x=\{f\in C(X):f(x)=0\}$, then this is a maximal ideal

   1. This part I’ll do low-tech for clarity. To show this is an ideal note that if $f,g\in M_x,h\in C(X)$

      $$(f+g)(x)=f(x)+g(x)=0$$

      $$(fh)(x)=f(x)h(x)=0$$

      and since $C(X)$ is commutative it follows that $M_x$ is a two-sided ideal.

   2. This part will be high-tech. On the other hand note that the map $e_x:C(X)\to ℂ$ defined by $e_x(f)=f(x)$ has exactly $M_x$ as its kernel. It follows then by commutativity of $C(X)$ and the fact that

      $$ℂ=Im{e}_x\cong C(X)/M_x$$

      is a field, and in particular a division ring, that $M_x$ is maximal.

   3. Of note is that this also applies to the ring of bounded complex-valued continuous functions

2. Let $M$ be a maximal ideal in $C(X)$ where $X$ is a compact Hausdorff space, then $M=M_x$ for some $x\in X$

   1. Note that it suffices to show that the set

      $$K_M=\bigcap _{f\in M}\{x\in X:f(x)=0\}$$

      is non-empty

   2. If not, then

      $$\bigcup _{f\in M}\{x\in X:f(x)\ne 0\}=X$$

      However, since $\{0\}$ is closed it follows that $\{x\in X:f(x)\ne 0\}$ is open and so the collection $\{x\in X:f(x)\ne 0\},f\in M$ forms an open cover. Then there exist $f_1,\mathrm{\dots },f_n\in M$ such that

      $$\bigcup _{i=1}^n\{x\in X:f_i(x)\ne 0\}=X$$

      Define $g\in M$ by

      $$g(x)=\sum _{i=1}^n{f}_i(x)\overline{f_i}(x)$$

      Since each term is real and non-negative and at least one term is positive at each point it follows that $g\ne 0$

   3. We therefore have that

      $$g\cdot \frac{1}{g}=1$$

      and so the ideal contains the identity element. This however implies that $M=C(X)$ which contradicts the definition of ideal. Thus $K_M$ is nonempty and so if $x\in K_M$ then $M=M_x$.

3. With all definitions as in problem 2 $M_y=M_x$ implies $x=y$

   1. We first note that a compact Hausdorff space is normal.

   2. Thus if $x\ne y$ Urysohn’s lemma provides a continuous function $f(z)$ such that $f(x)=1,f(y)=0$

   3. This implies that $f\in M_y,f\notin M_x$ which contradicts our definition.

   4. Thus $x=y$

4. Let $L^1({ℝ}^n)$ be the space of integrable complex-valued functions on ${ℝ}^n$ which forms a ring with convolution as the product, then the set

   $$M_{\xi }=\{f\in L^1({ℝ}^n):{\int }_{{ℝ}^n}f(x)e^{-2\pi ix\cdot \xi }𝑑x=0\}$$

   is a maximal ideal

   1. We note that this is almost a restatement of problem 1, but I found it interesting enough to include by itself and there are a few different technical points

   2. The Fourier transform is defined on $L^1({ℝ}^n)$ by $\widehat{f}(\xi )={\int }_{{ℝ}^n}f(x)e^{-2\pi ix\cdot \xi }𝑑x$ so that

      $$M_{\xi }=\{f\in L^1({ℝ}^n):\widehat{f}(\xi )=0\}$$

   3. It follows by the standard properties of the Fourier transform that

      $$\widehat{(f\star g)}=\widehat{f}\widehat{g}$$

      $$\widehat{(f+g)}=\widehat{f}+\widehat{g}$$

      and

      $$\widehat{}:L^1({ℝ}^n)\to BC({ℝ}^n)$$

      the latter being the space of bounded continuous functions. This implies that similarly to part a) of problem 1 $M_{\xi }=\{f\in L^1({ℝ}^n):\widehat{f}(\xi )=0\}$ is an ideal.

   4. Similarly to problem 1 we want to show that $ker(e_{\xi })=M_{\xi }$ and $L^1({ℝ}^n)/ker(e_{\xi })\cong ℂ$ for some ring homomorphism $e_{\xi }$.

   5. The evaluation map

      $$e_{\xi }(f)=\widehat{f}(\xi )$$

      is the correct mapping that we’re looking for. It is a ring homomorphism onto the complex numbers with $ker(e_{\xi })=M_{\xi }$ by the properties listed in part c). Furthermore, we may pick functions $f_z,z\in ℂ$ in $L^1({ℝ}^n)-ker(e_{\xi })$ such that

      $$e_{\xi }(f_z)=\widehat{f_z}(\xi )=z$$

      so that $M_x$ is maximal.