Regarding maximal ideal space

Question

Let $A$ be the set of all degree one polynomials of the form $a+bx$ with complex coefficients $a$ and $b$. Define the multiplication on $A$ by $$(a+bx)(c+dx)=ac+(ad+bc)x.$$ And let the norm $\|.\|$ on $A$ be $$\|(a+bx)\|=|a|+|b|.$$ Then $(A,\|.\|)$ is a commutative unital Banach algebra. They say that $(A,\|.\|)$ has only one maximal ideal, i.e the set of $a+bx$ such that $a=0$. Can anyone tell why? How exactly does one identify the maximal ideal space of $A$?

Answer 1

Another way of looking at this:

True Fact: If $\psi:\Bbb C[x]\to\Bbb C$ is a non-zero homomorphism then there exists $\alpha\in\Bbb C$ such that $\psi(p)=p(\alpha)$ for all $p$.

The True Fact is presumably well known, and if you know the True Fact the result follows:

Note first that the multiplication in $A$ can be described as follows: "Multiply $p$ and $q$ using ordinary polynomial multiplication and then discard the $x^2$ term in the product." That says that

$A\simeq\Bbb C[x]/<x^2>$.

Now say $I$ is a maximal ideal in $A$. The magic of Banach algebras says that $I$ is the kernel of some homomorphism onto $\Bbb C$ instead of just onto some field. If $j:\Bbb C[x]\to\Bbb C[x]/<x^2>$ is the canonical homomorphism then $\psi=j\circ\phi$ is a complex homomorphism of $\Bbb C[x]$. So there exists $\alpha\in\Bbb C$ with $\psi(p)=p(\alpha)$. But $j(x^2)=0$, hence $\psi(x^2)=0$, hence $\alpha=0$.

Ok, here's a proof of the True Fact. Although I should say that if it's algebra and I can figure it out then anyone who knows some algebra should have no problem:

Lemma: If $F$ is an algebraically closed field and $I$ is a proper ideal in $F[x]$ then there exists $\alpha\in F$ such that $p(\alpha)=0$ for all $p\in I$.

Proof: Assume wlog that $I\ne0$. Suppose that the polynomials in $I$ have no common zero. Let $p_1\in I$, $p_1\ne0$. Since $p_1$ has only finitely many zeroes there exist $p_2,\dots,p_n\in I$ such that $p_1,\dots,p_n$ have no common zero.

Since $F$ is algebraically closed this says that $p_1,\dots,p_n$ have no common factor; since $F[x]$ is a PID it follows that $I=F[x]$.

A stronger version of the True Fact follows immediately:\

Corollary: If $F$ is an algebraically closed and $I$ is a maximal ideal in $F[x]$ then there exists $\alpha\in F$ such that $I=\{p:p(\alpha)=0\}$.

Proof: The lemma says that there exists $\alpha$ such that $I\subset\{p:p(\alpha)=0\}$; since $I$ is maximal we must have $I=\{p:p(\alpha)=0\}$.

Answer 2

Normally, maximal ideals are not typically unique. They are just (proper) ideals that cannot be further extended without becoming the entire set.

Let $I$ be the proposed maximal ideal. You need to show, of course, that $I$ is an ideal. It's not hard to show that it's closed under addition, scalar multiplication, and multiplication by any other element of $A$.

It is also maximal, since if we take an ideal $J$ properly containing $I$, then some $a_0x + b_0$ exists in $J \setminus I$, i.e. where $b_0 \neq 0$. But then $-a_0x \in I$ and hence in $J$ too, so $a_0x + b_0 - a_0x = b_0 \in J$. Using the property of ideals, we also have $b_0^{-1} b_0 = 1 \in J$. Therefore, given any $ax + b \in A$, we have $(ax + b)1 \in J$, that is $J = A$.

Next, you need to show that $I$ is the only maximal ideal. Suppose $J$ is some arbitrary maximal ideal of $A$. Note that if $0x + b \in A$, and $b \neq 0$, then the above argument shows $J = A$. More generally, if we have $ax + b \in A$ with $b \neq 0$, then we have $$(ax + b)(-ax + b) = b^2 \in J,$$ with $b^2 \neq 0$, and hence $J = A$. Thus, if $ax + b \in J$, then $b = 0$, so $J \subseteq I$. From maximality of $J$, we conclude $J = I$.