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\centerline{\bf VIRTUAL LECTURE 5-29}
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Given any ring homomorphism $f:A\rightarrow B$ there is an induced map $f^*$ taking Spec$(B)$ to Spec$(A)$, which is continuous in the Zariski topologies on these spectra; this map can be used in conjunction with localization and passage to the quotient to relate the spectrum of a difficult ring to that of an easier one.  As an example, we will simultaneously work out the spectra of $\Bbb Z[x]$ and $k[x,y], k$ a field; these two very different rings exhibit many parallel features.  Letting $B$ be one of these rings and $A=\Bbb Z$ in the first case, $A = k[x]$ in the second, we begin by observing that any prime ideal of $B$ contracts to a prime ideal in $A$; if this is nonzero, then by passage to the quotient we are reduced to studying prime ideals of $\Bbb Z/(p)[x]$ in the first case (for $p$ a prime integer) and prime ideals of $k'[x]$ in the second (where $k'$ is a simple finite extension of $k$, obtained as the quotient of $k[x]/(q), q$ an irreducible monic polynomial over $k$.  In both cases the ring obtained is a PID, so we deduce immediately that {\sl prime ideals of $\Bbb Z[x]$ meeting $\Bbb Z$ nontrivially take the form $(p,f)$, where $p$ is a prime integer and $f\in\Bbb Z[x]$ is a monic polynomial whose reduction $\overline f$ mod $p$ is irreducible in $\Bbb Z/(p)[x]$}. Similarly, {\sl prime ideals in $k[x,y]$ meeting $k[x]$ nontrivially take the form $(q,q')$, where $q$ is monic irreducible in $k[x]$ and $q'$ is monic in $y$, its reduction $\overline q'$ mod $x$ being monic irreducible in $k[y]$}.  It remains to treat prime ideals of $Z[x]$ contracting to 0 in $\Bbb Z$, and similarly prime ideals of $k[x,y]$ contracting to 0 in $k[x]$; here localization comes to the fore, telling us that such ideals are in bijection to prime ideals of $\Bbb Q[x],k(x)[y]$, respectively.  As both of these rings are again PIDs, we deduce that {\sl the remaining nonzero prime ideals in $\Bbb Z[x]$ take the form $(f)$ for $f\in\Bbb Z[x]$ irreducible} (equivalently, $f\in\Bbb Z[x], f$ irreducible as a polynomial in $\Bbb Q[x]$, and $f$ primitive).  Once again a parallel result holds for $k[x,y]$.  

More generally, homomorphisms $A\rightarrow B$ behave especially nicely on the level of prime spectra if $B$ is flat as an $A$-module.  Recall from the fall quarter that this means (by definition) that tensoring with $B$ is an exact functor from $A$-modules to $A$-modules, and that $B$ is automatically flat over $A$ whenever $B$ is projective (in particular free) as an $A$-module.  Things are even nicer if $B$ is {\sl faithfully flat}; here (by one of several equivalent characterizations) the induced map from Spec$(B)$ to Spec$(A)$ is surjective, (equivalently, the extension of any maximal ideal of $A$ is proper in $B$).  There is however no requirement that the map from Spec$(B)$ to Spec$(A)$ be injective, and indeed $B$ can have a much larger spectrum than $A$ in this situation.  

Apart from flat extensions, the other kinds of especially nice ring extensions from this point of view are localizations.  The localization $B=A_S$ of a ring $A$ by a multiplicatively closed subset $S$ is such that every ideal of $B$ is the extension of an ideal of $A$.  Though in general different ideals of $A$ can have the same extension in $B$, this cannot happen for prime ideals:  {\sl there is a one-to-one inclusion-preserving correspondence between prime ideals of $B$ and prime ideals of $A$ not meeting $S$, given by extension from $A$ to $B$ and contraction from $B$ to $A$}.  (The same holds for primary ideals and was used to derive the uniqueness results in class for primary decompositions.)  This phenomenon helps explain the term \lq\lq localization"; in the special case where $S$ is the complement of a fixed prime ideal $P$, localization cuts out all prime ideals except those \lq\lq Zariski-close" to $P$ in the sense that they are contained in it.  (In a similar but more elementary manner, moding out by $P$ cuts out all primes except those {\sl containing} $P$.)  If $S$ is the complement of $P$, then in particular $A_S$ is a local ring, whose unique maximal ideal is generated by $P$.  

Recall also from last quarter that integral extensions of a ring also behave nicely in this respect.  We saw then (in our current language) that given an integral extension $A\subset B$, every prime ideal of $A$ is the contraction of some prime ideal of $B$; although there can be two primes in $B$ with the same contraction in $A$, there cannot be any inclusion relations between two such ideals.  In particular, the (Krull) dimensions of $A$ and $B$ always coincide.
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