Maximal Ideals in $L^1(\mathbb{R})$ Define $I_\xi = \{ f \in L^1(\mathbb{R}) : \hat{f}(\xi)=0 \}$. I have to prove that $I_\xi$ is a maximal ideal in $L^1(\mathbb{R})$. 
The following are my attempts at solution :
Attempt 1 :
Consider the multiplicative linear functional $\Phi$ defined by $\Phi(f)=\hat{f}(\xi)$. Then, $L^1(\mathbb{R})/\ker \Phi \simeq \mathbb{C}$. Thus, Ker $\Phi = I_\xi$ is a maximal ideal. Is there any problem with this solution?
However, I need to find a solution that uses only Fourier analysis. 
Attempt 2 :   I tried to prove this by contradiction, by assuming there exists a proper closed ideal $J \supset I$. Suppose $F \in J\setminus I$. Then, $\hat{F}(\xi) \neq 0$. If $\hat{F}$ is nowhere vanishing, Weiner Tauberian theorem will show $J=L^1(\mathbb{R}).$ If $\hat{F}(\alpha)=0$ for some $\alpha \in \mathbb{R}$, in order to use the Weiner Tauberian argument, I need to find $g \in J(\text{ or } I)$ such that $\hat{g}(\alpha) \neq 0$. This is the point where I am stuck. Any hint would be much appreciated
 A: The multiplication operation on $L^1(\mathbb{R})$ is convolution. If you choose $f \in L^1$ such that $f \ne 0$ and $\hat{f}(\xi)=0$ for a fixed, given $\xi\in\mathcal{R}$, then $h=f\star g = g\star f$ also has the property that $\hat{g}(\xi)=0$ (for this you need the Fourier transform and $\widehat{f\star g}=C\hat{f}\hat{g}$ where $C$ is a constant. And, if $f,g\in L^1$ satisfy $\hat{f}(\xi)=0$, $\hat{g}(\xi)=0$, then $(\widehat{f+g})(\xi)=0$. So it's not hard to see that
$$
              \mathscr{I}_{\xi}=\{ f\in L^1 : \hat{f}(\xi)=0 \}
$$
is an ideal of the convolution algebra $(L^1,\star)$. If $g \in L^1\setminus\mathscr{I}_{\xi}$, then $g$ can be normalized so that $\hat{g}(\xi)=1$. For any $h\in L^1$, you can write $h=\{h-\hat{h}(\xi)g\}+\hat{h}(\xi)g$, and $h-\hat{h}(\xi)g \in \mathscr{I}_{\xi}$. It follows that if $h,k\in L^1$, then
$$
        h\star k - \hat{h}(\xi)\hat{k}(\xi)g\star g \in \mathscr{I}_{\xi}, \\
        h\star k - \hat{h}(\xi)\hat{k}(\xi)g \in \mathscr{I}_{\xi}.
$$
So the algebra homomorphism $\omega_{\xi} : L^1 \rightarrow L^1/\mathscr{I}_{\xi}$ is $\omega_{\xi}(h)=\hat{h}(\xi)[g]$.
