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For a commutative unital Banach algebra $A,$ and $x\in A,$ we have $\lambda \in \sigma_A(x)$ if and only if $\phi(x) = \lambda$ for some algebra homomorphism $\phi:A \to \mathbb C.$ The set of all such homomorphisms, denoted $\Phi_A,$ is called the spectrum of $A.$

In algebraic geometry, the spectrum of $A$ is simply the set of all prime ideals of $A$ with additional structure. However, if we just consider algebra homomorphism $\phi:A \to \mathbb C,$ then we are just considering the maximal ideal, not the prime ideals.

The question is: are these two notions of spectrums related? Why prime ideals need not be considered, just the maximal ideals, on this occasion?

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    $\begingroup$ It is also rather common for a "user's approach" to algebraic geometry to identify an affine variety $X$ over an algebraically closed field $k$ with its set of $k$-valued points, i.e., with the set of maximal ideals in $k[X]$. (For example, Borel - Linear algebraic groups takes this approach.) $\endgroup$
    – LSpice
    Commented Apr 7 at 19:02
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    $\begingroup$ The maximal ideals are the Zariski closed points of the prime spectrum. $\endgroup$
    – JJJ
    Commented Apr 7 at 19:06
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    $\begingroup$ In algebraic geometry, the maximal spectrum is not of much use for a local ring, and fails to distinguish between local rings and fields, so the prime spectrum is used (there are other good reasons as well). The reason we don't use the prime spectrum in Banach algebra theory is that in general there are a lot of weird prime ideals in $C(X)$ that don't have a geometric interpretation. Examples can be found in Gillman and Jerison's Rings of Continuous Functions. (Of course, if anyone knows a use for these ideals, other than as counterexamples, I'd be glad to hear it.) $\endgroup$
    – Robert Furber
    Commented Apr 7 at 20:26
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    $\begingroup$ You might like the exposition in Chapter 7 of Palmer's book books.google.ca/books?id=3k0lKh6QY-QC&pg=PA619 $\endgroup$
    – Onur Oktay
    Commented Apr 9 at 9:40

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