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Refer to localization of rings in Lang's Algebra p. 108. In particular, let $A$ be a commutative ring, $S$ a subset of $A$ that is a submonoid of the multiplicative monoid structure of $A$, containing the identity. Then $S^{-1}A$ is the set of elements $\frac{a}{s}$ with $a \in A$ and $s \in S$. We have $\frac{a_1}{s_1}=\frac{a_2}{s_2}$ if there exists $\bar{s} \in S$ such that $\bar{s}(s_2a_1 - s_1 a_2)=0$. Now consider the injection of $A$ into $S^{-1}A$ given by $x \mapsto \frac{x}{1}$. This is a ring homomorphism. Let $J$ be an ideal of $A$. Ring homomorphisms preserve ideals and so the injection of $J$ into $S^{-1}A$, call it $J^{*}$, will be an ideal. My problem is that i don't see why for any $\frac{a}{s} \in S^{-1}A$ and any $\frac{j}{1} \in J^{*}$ we will have that $\frac{aj}{s}$ will be in $J^{*}$, i.e. why there must exist an element $\kappa \in J$ such that $\frac{aj}{s}=\frac{\kappa}{1}$.

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It's not a good idea to call the natural map from $A$ to $S^{-1}A$ an "injection", since it isn't actually injective in general. – Chris Eagle Oct 30 '11 at 19:43
thanks, you are right. – Manos Oct 30 '11 at 21:37
up vote 5 down vote accepted

Just a note, since Chris has already cleared things up. When people talk about moving ideals from $A$ to $S^{-1}A$, or more generally from $A$ to $B$ using a homomorphism $f\colon A \to B$, they really mean: take an ideal $\mathfrak{a}$ of $A$ and form the ideal $f(\mathfrak{a})B$ of $B$, i.e. the ideal of $B$ generated by the set $f(\mathfrak{a})$. This is called the extension of the ideal and is sometimes denoted by $\mathfrak{a}^e$, or $S^{-1}\mathfrak{a}$ in the case of localization.

For localizations the story of how ideals extend (and contract) is very interesting. See Proposition 3.11 of Atiyah-Macdonald or Proposition 6.4 of Milne's notes for a discussion of this.

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What exactly is the set $f()$? It is the image of the ideal of $A$ under $f$? – Manos Oct 30 '11 at 21:28
@Manos Exactly. $f(\mathfrak{a}) = \{f(a) : a \in \mathfrak{a}\}$. – Dylan Moreland Oct 30 '11 at 21:45
and something else: let $p$ be a prime ideal of $A$ and $M, N$ $A$-modules. Let $M_p, N_p$ be the localized modules at $p$ and let $f:M_p \rightarrow N_p$ be an $S^{-1}A$-module homomorphism, $S$ being the complement of $p$ in $A$. Then how does this $f$ induce a homomorphism $M \rightarrow N$? – Manos Oct 30 '11 at 23:16
I will post this as a new question... – Manos Oct 30 '11 at 23:31
@Manos It's not clear to me that what you ask is always possible. – Dylan Moreland Oct 31 '11 at 0:09

The claim is false: ring homomorphisms do not preserve ideals. For example, take $A=\mathbb{Z}$ and $S=\mathbb{Z} \setminus \{ 0 \}$, so that $S^{-1}A = \mathbb{Q}$. Then $\mathbb{Z}$ is of course an ideal of itself, but its image is not an ideal of $\mathbb{Q}$.

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Only prime ideals which are disjoint from S have 1 to 1 corresponds with prime ideals in $S^{-1}A$. There are no general mapping between ideals. BTW, I got problem 3.

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