Taking Fourier transforms, you get the action by multiplication. (Really, $x$ acts by multiplying with $e^{ix}$.) The closed invariant subspaces will thus be of the form ${fin L^2(mathbb{R})colonhat f|_E=0}$ for measurable sets $E$.
But you were asking for closed irreducible subspaces: There aren't any, because the Lebesgue measure has no atoms.
One answer to your question is to remain within the realm of the continuum, and consider what happens when the Continuum Hypothesis fails. In this case, there are cardinals κ below the continuum c = |R|, and one naturally inquires whether these cardinals behave more like aleph0, or more like the continuum, with respect to measure and additivity. For example, we may still want to inquire about our favorite measures, but with uncountable cardinals.
The answer is saturated with set-theoretic independence. For example, it is known to be consistent with the axioms of set theory that Lebesgue measure can be better than countably additive! Under Martin's Axiom (MA), when the Continuum Hypothesis fails, then
the union of κ many measure zero sets remains measure zero, for any κ below the continuum. It follows from this that Lebesgue measure is literally ≤κ additive, in the sense that the measure of the union of κ many disjoint sets is the sum of their individual measures (since only countably many of them can have positive measure). This includes the case of uncountable κ, and so goes strictly beyond countable additivity. There are similar results concerning the additivity of the ideal of meager sets.
Indeed, there is a rich subject investigating this called cardinal characteristics of the continuum. I discussed some of the concepts in this MO answer. The point of this subject is to investigate exactly how the dichotomy between countable and continuum plays out in situations when CH fails. Researchers in this area define a number of cardinal invariants, such as:
The bounding number b is the size of the smallest unbounded family of functions from ω to ω. There is no function that bounds every member of the family.
The dominating number d is the size of the smallest dominating family of functions ω to ω. Every function is dominated by a member of the family.
The additivity number for measure is the smallest number of measure zero sets whose union is not measure zero.
The covering number for measure is the smallest number of measure zero sets whose union is all of R.
The uniformity number for measure is the size of the smallest non-measure zero set.
The cofinality number for measure is the smallest size of a family of measure zero sets, such that every measure zero set is contained in one of them.
Each of these numbers is ω1 under the Continuum Hypothesis, and this expresses the preoccupation with ω that you mention in your question. However, when CH fails, then one cannot prove that any of these numbers is equal to another. Each of them expresses a fundamental characteristic of the continuum, and there are models of set theory distinguishing any two of them (and distinguishing them from ω1 and the continuum as well).
One can define similar numbers using the ideal of meager sets in place of the ideal of measure zero sets, and the relationships between all these cardinal characteristics are precisely expressed by Cichon's diagram. In particular, no two of them are provably equal, and there are models of set theory exhibiting wide varieties of possible relationships. There are dozens of other cardinal characteristics, whose relationships are the focus of intense study by set theorists working in this area. The main tool for separating these cardinal characteristics is the method of forcing and especially iterated forcing.
Another answer to your question is to go well beyond the continuum, and perhaps this is what you are really asking about. The analogue of the theory of cardinal characteristics of the continuum has been carried out for arbitrary cardinals κ, and again, the situation is saturated with independence results, proved by forcing.
For certain types of questions, however, it is interesting to note that there is no intermediate possibility between the countable and large cardinals. For example,
an ultrafilter on a set can be viewed as an ω-complete 2-valued measure on subsets of ω (ω complete = additive for unions of size less than ω = finitely additive). Can one have an Aleph1 additive ultrafilter on a set? Well, if F is such an ultrafilter, then it must also be Aleph2 additive, Aleph3 additive, and so on, for quite a long way, up to the least measurable cardinal. That is, it is just not possible to have a δ-additive ultrafilter, when δ is uncountable, unless it is also additive up to measurable cardinal. Other similar phenomenon surround the question: does every κ additive filter extend to a κ additive ultrafilter on a set? This is true for κ = ω, since this just amounts to being finitely additive, which is what being a filter means. But for uncountable κ, it is equivalent to the assertion that κ is strongly compact, which is very high in the large cardinal hierarchy.
With all this unanimous enthusiasm, I can't help but add a cautionary note. I will say, however, that what I'm about to say applies to anyone of any age trying to get a Ph.D. and pursue a career as an academic mathematician.
If you think you want a Ph.D. in mathematics, you should first try your best to talk yourself out of it. It's a little like aspiring to be a pro athlete. Even under the best of circumstances, the chances are too high that you'll end up in a not-very-well-paying job in a not-very-attractive geographic location. Or, if you insist on living in, say, New York, you may end up teaching as an adjunct at several different places.
Someone with your mathematical talents and skills can often find much more rewarding careers elsewhere.
You should pursue the Ph.D. only if you love learning, doing, and teaching mathematics so much that you can't bear the thought of doing anything else, so you're willing to live with the consequences of trying to make a living with one. Or you have an exit strategy should things not work out.
Having said all that, I have a story. When I was at Rice in the mid 80's, a guy in his 40's or 50's came to the math department and told us he really wanted to become a college math teacher. He had always loved math but went into sales(!) and had a very successful career. With enough money stashed away, he wanted to switch to a career in math. To put it mildly, we were really skeptical, mostly because he had the overly cheery outgoing personality of a salesman and therefore was completely unlike anyone else in the math department. It was unthinkable that someone like that could be serious about math. Anyway, we warned him that his goal was probably unrealistic but he was welcome to try taking our undergraduate math curriculum to prepare. Not surprisingly, he found this rather painful, but eventually to our amazement he started to do well in our courses, including all the proofs in analysis. By the end, we told the guy that we thought he really had a shot at getting a Ph.D. and have a modest career as a college math teacher. He thanked us but told us that he had changed his mind. As much as he loved doing the math, it was a solitary struggle and took too much of his time away from his family and friends. In the end, he chose them over a career in math (which of course was a rather shocking choice to us).
So if you really want to do math and can afford to live with the consequences, by all means go for it.
A prime ideal of $S^{-1}R[X]$ is the extension of a unique prime ideal of $R$, so that the morphism $Spec(S^{-1}R[X])to Spec(R)$ is a bijection, and even an homeomorphism. All the extensions of residual fields induced are pure transcendental of transcendence degre $1$.
As an example, if you look at the case $R=mathbb{Z}$, the morphism of schemes you get "puts in family" the extensions of fields $mathbb{F}_phookrightarrowmathbb{F}_p(X)$.
The answer is no. Here is a counter-example
$$left( begin{matrix} 0 & -5 \\ 1 & 0 end{matrix} right) quad mbox{and} quad left( begin{matrix} -1 & -3 \\ 2 & 1 end{matrix} right).$$
Both of these matrices have characteristic polynomial $x^2+5$. For $p neq 2$, $5$, this polynomial has no repeated factors so any matrices with this polynomial are similar. By the same argument, they are similar over $mathbb{Q}$. By brute force computation, they are also similar at $2$ and $5$.
However, they are not similar over $mathbb{Z}$. Consider $mathbb{Z}^2$ as a module for $mathbb{Z}[t]$ where $t$ acts by one of the two matrices above. Both of these matrices square to $-5$, so these are in fact $mathbb{Z}[sqrt{-5}]$-modules. If the matrices were similar, the similarity would give an isomorphism of $mathbb{Z}[sqrt{-5}]$-modules. But these are not isomorphic: the former is free on one generator while the latter is isomorphic to the ideal $langle 2, 1+ sqrt{-5} rangle$.
In general, the way to classify similarity of matrices over $mathbb{Z}$ is the following: If the matrices do not have the same characteristic polynomial over $mathbb{Q}$, they are not similar. If they do, let $f$ be the characteristic polynomial and let $R=mathbb{Z}[t]/f(t)$. Then your matrices give $R$-modules, and the matrices are similar if and only if the $R$-modules are isomorphic. If $R$ is the ring of integers of a number field, then $R$-modules are classified by the ideal class group. In general, they are related to the ideal class group, but there are various correction factors related to how $R$ fails to be the ring of integers of its fraction field (or how it fails to be a domain at all). I don't know the details here.
To add to Franz's nice answer:
Let $mathcal{C}$ be the collection of groups isomorphic to the direct sum of a free abelian group of countable rank with a finite abelian group.
In the case where $L/K$ is a nontrivial finite separable extension of global fields, each of the following groups is in $mathcal{C}$ (and the first two are free):
1) The group of fractional ideals of $K$ (or divisors in the function field case)
2) The group of principal fractional ideals of $K$
3) $K^times$
4) $L^times/K^times$
Proof: The first three can be proved in succession by using infinitude of primes, finiteness of class groups, and the Dirichlet unit theorem.
4) As suggested by t3suji and Franz, Chebotarev shows that the rank is infinite. On the other hand, the following trick shows that $L^times/K^times$ is a subgroup of a group in $mathcal{C}$ (and hence in $mathcal{C}$ itself): Replace $L$ by its Galois closure. Let $sigma_1,ldots,sigma_d$ be the elements of $operatorname{Gal}(L/K)$. Then
$$x mapsto (sigma_1(x)/x,ldots,sigma_d(x)/x)$$
injects $L^times/K^times$ into $L^times times cdots times L^times$,
which is in $mathcal{C}$.
A Hecke algebra describes the most reasonable way to convolve functions or measures on a homogeneous space. Suppose that you have seen the definition of convolution of functions on a vector space, or on a discrete group --- the latter is just the group algebra of the group or some completion. Then how could you reasonably define convolution on a sphere? There is no rotationally invariant way to convolve a general $f$ with a general $g$. However, if $f$ is symmetric around a reference point, say the north pole, then you can define the convolution $f * g$, even if $g$ is arbitrary.
This is the basic idea of the Hecke algebra. The $(n-1)$-sphere is the homogeneous space $SO(n)/SO(n-1)$. A function $g$ on the sphere is a function on left cosets. A function $f$ on the sphere which is symmetric about a reference point is a function on double cosets. If $H subseteq G$ is any pair of compact groups, if $f$ is any continuous function on $Hbackslash G/H$, and if $g$ is any continuous function on $G/H$, then their product in the continuous group algebra is well-defined on $G/H$. The functions on double cosets make an algebra, the Hecke algebra, and the functions on left cosets are a bimodule of the Hecke algebra and the parental group $G$.
It is important for the same reasons that any other kind of convolution is important.
A particular case studied by Hecke Iwahori and others from before quantum algebra was the finite group $GL(n,q)$ and the upper triangular subgroup $B$. This is "the" Hecke algebra; it turns out that it is one algebra with a parameter $q$. Or as Ben says, this generalizes to the Iwahori-Hecke algebra of an algebraic group $G$ with a Borel subgroup $B$.
The other place that the Hecke algebra arises is as an interesting deformation of the symmetric group, or rather as a deformation of its group algebra. It has a parameter $q$ and you obtain the symmetric group when $q=1$. As I said, it is also the Hecke algebra of $GL(n,q)/GL(n,q)^+$, where $B = GL(n,q)^+$ is the Borel subgroup of upper-triangular matrices (all of them, not just the unipotent ones). There is a second motivation for the Hecke algebra that I should have mentioned: It immediately gives you a representation of the braid group, and this representation reasonably quickly leads to the Jones polynomial and even the HOMFLY polynomial.
When the Jones and HOMFLY were first discovered, it was simply a remark that the braid group representation was through the same Hecke algebra as the convolutional Hecke algebra for $GL(n,q)/B$ (or equivalently $SL(n,q)/B$). Even so, it's a really good question to confirm this "coincidence", as Arminius asks in the comment. Particularly because it is now a fundamental and useful relation and not a coincidence at all. As Ben explains in his blog post, the first model of the Hecke algebra is important for the categorification of the second model.
The coset space of $GL(n,q)/B$ consists of flags in $mathbb{F}_q^n$, and you can see these more easily using projective geometry. When $n=2$, there is an identity double coset 1 and another double coset $T$. A flag is just a point in $mathbb{P}^1$, and the action of $T$ is to replace the point by the formal sum of the other $q$ points. Thus you immediately get $T^2 = (q-1)T + q.$ When $n=3$, a flag is a point and a line containing it in $mathbb{P}^2$. The two smallest double cosets other than the identity are $T_1$ and $T_2$. $T_1$ acts by moving the point in the line; $T_2$ acts by moving the line containing the point. A little geometry then gives you that $T_1T_2T_1$ and $T_2T_1T_2$ both yield one copy of the largest double coset and nothing else. Thus they are equal; this is the braid relation of the Hecke algebra. When $n ge 3$, the Hecke algebra is given by these same local relations, which must still hold.