Computing the spectrum of an algebra Definitions used:


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*Character of an $\mathbb{R}$-algebra A: a linear function $\chi_{x}: A \rightarrow \mathbb{R}, \chi_{x}: f \mapsto f(x)$ for $x \in \mathbb{R}$ so that $\chi(uv) = \chi(u)\chi(v)$.

*Spectrum of A: The collection X = $\{ \; \chi_{x} : x \in \mathbb{R} \; \}$ of characters of A.

*Topological spectrum: The spectrum of A equipped with the smallest topology so that all evaluation functions $E_{a} : X \rightarrow \mathbb{R}, E_{a}: \chi \mapsto \chi(a)$ are continuous.


Hi, I need to compute the spectrum of characters of $\mathbb{R}[t]$, the algebra of real polynomials in one variable, in order to prove that the topological spectrum is homeomorphic to $\mathbb{R}$. I could surely find some characters, but how should I go about constructing the entire topological spectrum? In general, what could be things to try first, as I have several exercises of this form? Thanks in advance. 
 A: You have defined the spectrum to be the evaluation maps of points in $\Bbb R$, so you have a natural bijection $\varphi: X\to \Bbb R$ via $\chi_x\mapsto x$. You want to show that this is a homeomorphism.
The topology of $X$ has as basis $U_\epsilon(f;y)=\{\chi_x\mid \chi_x(f)=f(x)\in B_\epsilon(y)\}=\{\chi_x\mid x\in f^{-1}(B_\epsilon(y))\}$ and then $\varphi(U_\epsilon(f;y) )= f^{-1}(B_\epsilon(y))$ which is an open subset of $\Bbb R$, so the map $\varphi$ is open.
Now $\varphi^{-1}(B_\epsilon(y))=\{\chi_x\mid x\in B_\epsilon(y)\}=U_\epsilon(\mathrm{id},y)$ is also open and $\varphi$ is continuous.
So its a homeomorphism.
Some further words:
Characters are defined to be non-zero algebra morphisms $A\to\Bbb K$ where $\Bbb K$ is the base field. If $A$ is a "function algebra" on a space $X$ then the characters do not need to correspond to evaluation maps, but the evaluation maps certainly are characters (cf this question for example). In the case of a complex Banach algebra characters correspond to maximal ideals in a one-to-one fashion, so this is also a way of exploring the spectrum.
