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In the book Geometry of Schemes, written by David Eisenbud & Joe Harris, at the start of chapter one in the section Schemes as Sets, the authors introduce elements of $R$ as functions. I can't understand their meanings there! If anyone knows how these functions are defined, please describe what functions that correspond to elements of $R$ are.

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2 Answers 2

If $R$ is a commutative ring one associates to it a set, the spectrum $Spec(R)$ of that ring.
The elements of that spectrum are the prime ideals $\mathfrak p\subset R$.
Now, given an element $r\in R$ one associates to it a function $\hat r $ defined on $Spec(R)$ (the Gel'fand transform of $r$), whose value at $\mathfrak p\subset R$ is the residue class $\hat r(\mathfrak p)=\text {class} (r)\in R/\mathfrak p\subset \kappa (\mathfrak p) =Frac(R/\mathfrak p)$.

This is an amazing construction.
Many rings are naturally given as functions on some structured set: think of continuous functions on a topological space or smooth functions on a manifold or holomorphic finctions on a holomorphic variety or...
The fantastic idea, due to Gel'fand for Banach algebras and to Grothendieck for general rings, is that you can essentially force every ring to be a ring of functions on some set, namely $Spec(R)$, extracted from $R$ itself.
Of course there are technical problems: the association $r\mapsto \hat r$ is not always injective , so that the ring $R$ is not always exactly a ring of functions, but this idea of trying to consider any ring as a ring of functions is really extraordinary and extremely useful.

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Also, $\hat{r}$ may not be a function since the fields of fractions $\operatorname{Frac}(R/\mathfrak{p})$ need not be the same at different points $\mathfrak{p}\in\operatorname{Spec}(R)$. –  Javier Álvarez Nov 15 '12 at 15:14
    
Dear @Javier, $\hat r$ is always a function. But its codomain is admittedly a bit weird: it is the union $\cup \mathfrak p \text {Frac} (R/\mathfrak p)$ of all the residual fields of the primes making up $Spec(R)$. –  Georges Elencwajg Nov 15 '12 at 19:01
    
yes indeed, any one-image assignment to each point is a function by considering the union of the possible codomains. I should have said something like "traditional function", as I was thinking in the motivation where $R$ is the coordinate ring of an algebraic variety over a field $k$, so $r$ takes values in $k$ at any closed point. You are right that the striking contribution is considering such "general functions" beyond those taking value in something like a number field, so that $\hat r$ contains much more information, i.e. all its restrictions to any subscheme. –  Javier Álvarez Nov 15 '12 at 20:19
    
I understand what you say in your last paragraph as: the "general function" $\hat r$ is not necesssarily determined by the values of it at every point (closed or generic). A silly example: for $k$ a field, then $R=k[x]/(x^2)$ has $\operatorname{Spec}(R)=(x)=:\mathfrak{p}_x$, thus $x\in R$ and although $x\neq 0$, it has $\hat x (\mathfrak{p}_x)=0$; in general there can be cases where $\hat r$ does not determine $r$ by its values at every point, so calling $r$ a function because of $\hat r$ must be done with care. (That is why in Eisenbud-Harris' book they emphasize quotation marks "function"). –  Javier Álvarez Nov 15 '12 at 20:34

In the book Geometry of Schemes, written by David Eisenbud \& Joe Harris, at the start of chapter one in the section Schemes as Sets, the authors introduce elements of R as functions. we will describe what functions that correspond to elements of R are.

In The Name Of Allah The Merciful The Compassionate

Let us recall some Theorems from fundamental Algebras

Theorem: Assume $R$ is a commutative ring with an identity element, then "$P$ is a prime ideal of $R$ $\Longleftrightarrow$ $\frac{R}{P}$ be an integral domain".\footnote{"Algebra" written by Dr. Mohammadi Hasan Abadi}

having identity element was needed for $\frac{R}{P}$ will be an integral domain, because we defined a ring $R$ is an integral domain if it will be commutative with identity element and has no non-zero zero divisor. But if you look at the proof of this theorem it is not needed to implies from $P$ is a prime ideal of $R$ that $\frac{R}{P}$ has no non-zero zero divisor and its convert.

Theorem: Assume $R$ is a commutative ring with at least two elements and has no non-zero zero divisors. Then the relation $\thicksim$ on $\{(a,b)\in R\times R\:|\:b\neq 0\}$ with definition "$(a,b)\thicksim(c,d)\Longleftrightarrow ad=bc$" is an equivalence relation on the set of quotients of $R$.\footnote{"Algebra" written by Dr. Mohammadi Hasan Abadi, page 299}

Theorem: The set of equivalence classes introduced in previous theorem with operations $+$ and $\cdot$ that are defined below is a field.\footnote{"Algebra" written by Dr. Mohammadi Hasan Abadi, page 301}

$\{ +:F\times F\longrightarrow F \\ (a,b)+(c,d)=(ad+bc,bd)$

$ \{\begin{array}{c} \cdot:F\times F\longrightarrow F \\ (a,b)\cdot(c,d)=(ac,bd) \end{array}$

Theorem: Assume $R$ is a commutative ring with at least two elements and has no non-zero zero divisor and let $F$ be its quotients field. Then R is embeddable in $F$.\footnote{"Algebra" written by Dr. Mohammadi Hasan Abadi, page 302}

Note that the identity of quotient field is equivalence class contains $(a,a)$ that $a\in R-\{0\}$ is arbitrary. And note that, embedding function that is mentioned is

$\{\begin{array}{c} f:R\longrightarrow F \\ f(a)=[(ab,b)]_{\thicksim}\quad ;b\in R-\{0\} \end{array}$

Be care that by definition of $\thicksim$, $[(ab,b)]_{\thicksim}$ is independent from $b$ and only depends on $a$.

Now if $R$ be commutative ring for every prime ideal of $R$ like $P$ we have $\frac{R}{P}$ is commutative ring with no non-zero zero divisors, so by mentioned theorems, $F_{P}$ the set of equivalence classes of relation $\thicksim$ on $\{(a+P,b+P)\in \frac{R}{P}\times \frac{R}{P}\:|\:b+P\neq P\}$ with mentioned operators $+$ and $\cdot$ will be a field that $\frac{R}{P}$ will be embedded in it with following function.

$\{\begin{array}{c} f_{P}:\frac{R}{P}\longrightarrow F_{P} \\ f_{P}(r+P)=\frac{(r+P)(s+P)}{s+P}\quad ;s+P\in \frac{R}{P}-\{P\} \end{array}$

Now the main concept of those authors will be shined as below:

For every $r\in R$ define; $\{\begin{array}{c} f_{r}:Spec(R)\longrightarrow \bigcup_{P\in Spec(R)}F_{P} \\ f_{r}(P)=\big(f_{P}o\pi_{P}\big)(r)=\frac{(r+P)(s+P)}{s+P}\quad ; s\in P \end{array}$

$\{\begin{array}{c} \pi_{P}:R\longrightarrow \frac{R}{P} \\ \pi_{P}(r)=r+P \end{array}$

For every one of elements of $R$ like $r$, $f_{r}$ is a function because if $P_{1},P_{2}\in Spec(R)$ and $P_{1}=P_{2}$ then it is like you are denoting a prime ideal of $R$ with two nomads, so $\frac{R}{P_{1}}=\frac{R}{P_{2}}$ and $F_{P_{1}}=F_{P_{2}}$ and for every $s\notin P_{1}$ we have $s\notin P_{2}$ and conversely and at the end $\frac{(r+P_{1})(s+P_{1})}{s+P_{1}}=\frac{(r+P_{2})(s+P_{2})}{s+P_{2}}$ that means $f_{r}(P_{1})=f_{r}(P_{2})$.

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Some notations are meaningless here for others so, it is better to avoid them. +1 :) –  B. S. Nov 16 '12 at 20:16

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