Archive for April, 2009

Prime ideals and rings of fractions

10 April, 2009

The concept of prime ideal in a commutative ring R with 1 is one of several natural generalisations of the concept of prime integer number.

An ideal I\subset R is called prime if for any u,v\in R the following holds:
uv\in I implies that at least one of these elements, u and v, is in I.

E.g. the principal ideal (p)\subset \mathbb{Z} is prime if and only if p is prime.

A maximal ideal I\subset R is prime.

Indeed, let uv\in I. Assume that u\not\in I. We need to show that then v\in I. If this were not the case then u+I and v+I are two non-zero elements in the ring R/I such that (u+I)(v+I)\subseteq I. But this is not possible, as R/I is a field. Thus v\in I, as claimed.

Analysing this proof, one can easily see that

If I\subset R is prime then R/I has no zero-divisors, i.e. it is an integral domain.

Further important property of prime ideals is that they are radical, i.e.
I=\sqrt{I}, where the radical \sqrt{I} of the ideal I ideal is \sqrt{I}=\{x\in R\mid \exists k\geq 1 : x^k\in I\}. Indeed, x^k\in I implies that either x or x^{k-1} is in I, and we derive x\in I by applying this reduction.

Yet another interesting observation is that R\setminus I is multiplicatively closed.

Rings of fractions

A subset S\subset R is called multiplicatively closed (or multiplicative) if 0\not\in S, 1\in S, and uv\in S for any u,v\in S.

Given a multiplicatively closed set S\subset R, one defines
a relation on R\times S, as follows:

{\displaystyle (y,t)\equiv (x,s)\quad\text{iff there exists }u\in S: uys=uxt.}

It is not hard to show that \equiv is an equivalence relation.
To simplify the notation, write its equivalence class with representative (x,s) as \frac{x}{s}. We define

S^{-1}R=(R\times S)/\equiv is the ring of fractions of R w.r.t. S, with addition and multiplication given by the rules
\frac{x}{s}+\frac{y}{t}=\frac{xt+ys}{st}, \frac{x}{s}\cdot\frac{y}{t}=\frac{xy}{st}.

It is easy to check that this is well-defined. We also have

\phi: R\to S^{-1}R, so that \phi(x)=\frac{x}{1}, is a ring homomorphism.

Note that \phi need not be injective, i.e. \phi(R)\cong R need not hold. Indeed, if x\in R is a zero-divisor such that xu=0 for u\in S then (x,1)\equiv (0,1), and so \phi(x)=\frac{x}{1}=\frac{0}{1}=0_{S^{-1}R}.

The most well-known example is the case R being an integral domain, and S=R\setminus\{0\}. Then S^{-1}R is a field, called the field of fractions of R.

Examples:

  • \mathbb{Q} is the field of fractions of \mathbb{Z}
  • for the ring of polynomials \mathbb{F}[T] over a field \mathbb{F} the field \mathbb{F}(T) is the field of rational functions over \mathbb{F} .
  • Let x\in R be non-nilpotent. Then S=\{1,x,x^2,x^3,\dots\} isa multiplicative set. Moreover, then S^{-1}R\cong R[T]/(Tx-1) (it is not completely trivial to prove this, though). Intuitively, we make the variable T behave like the inverse of x, as Tx=1 in this ring.

Now let us look at the case S=R\setminus I, for I a nonzero prime ideal. In this case S^{-1}R is denoted by R_I and called the localisation of R at I.

Example: Let \mathbb{F}[T] be the ring of polynomials over a field \mathbb{F}, and a\in \mathbb{F}. Then I=(T-a) is prime, and R_I=\mathbb{F}[T]_{(T-a)} is equal to \{\frac{f}{g}\in \mathbb{F}(T)\mid (T-a)\not|\, g \}.

The ring R_I has unique maximal ideal IR_I.

It suffices to show that \frac{x}{s} is invertible in R_I iff {x}\not\in I. Indeed, if x\in S then \frac{x}{s}\frac{s}{x}=1. On the other hand, if \frac{x}{s}\frac{y}{t}=1 then there exists u\in S such that uxy=ust. Thus uxy\not\in I, and so x\not\in I (if it was, uxy would be in I, as I is an ideal).

Dual spaces

5 April, 2009

There is one glaring omission in our Linear Algebra curriculum – it avoids talking about the dual space of a vector space. This makes talking about relationship between subspaces and equations that define them exceedingly difficult. Better late than never, so here it comes.

Let V be a vector space over a field \mathbb{F}. Denote by V^* the set of linear functions V\to \mathbb{F}.

Examples
Let V=C[a,b], the space of continuous functions on [a,b]. Then the function \int: V\to V given by f\mapsto \int_a^b f(x) dx is linear on V.

Let V=\mathbb{R}[x] be the vector space of polynomials with real coefficients.
Then the function V\to V given by f\mapsto \frac{df}{dx}(0) is linear on V.

Note that as f\in V^* is linear, one has f(\alpha v)=\alpha f(v) for any v\in V, \alpha\in \mathbb{F}. Thus we have m_\alpha:V^*\to V^* defined by m_\alpha(f)(v)=f(\alpha v) so that m_\alpha(m_\beta(f))=(m_\alpha m_\beta)(f). To simplify notation, we will write \alpha f instead m_\alpha(f). As well, we can define (f+g)(v)=f(v)+g(v) for any f,g\in V^*, and more generally, (\alpha f+\beta g)(v)=\alpha f(v)+\beta g(v). And there is the zero function 0(v)=0 for any v\in V. Thus we have all the ingredients of a vector space, as can be easily checked.

V^* is a vector space over \mathbb{F}. It is called the dual space of V.

So far, we haven’t used the linearity of our functions at all (we actually did not need the fact that \alpha f(v)=f(\alpha v)). Indeed, any closed under addition and multiplication set of functions V\to V would form a vector space.
What makes the dual space so special is that to define f\in V^* it suffices to define f(e_i) on a basis \{e_i\} of V. Indeed, f(\sum_i \alpha_i e_i)=\sum_i \alpha_i f(e_i), so we can compute f(v) for any v=\sum_i \alpha_i e_i, once we know the f(e_i)‘s.

Thus for a finite-dimensional vector space V one sees a (dependent upon the choice of a basis in V) bijection between V and V^*. This bijection, that is even an isomorphism of vector spaces, is defined by the dual basis of V^* given by coordinate functions x_i=\epsilon_i(x), where x_i‘s are the coefficients of x\in V is the decomposition of x in the basis \{ e_i\}.

Finite-dimensionality is crucial here. E.g. let us consider the vector space of polynomials \mathbb{Z}[x]. It is a countable space: one can view it as the set of infinite 0-1 strings, with only finitely many 1’s occurring in each string. On the other hand, V^* can be viewed as the set of all the infinite 0-1 strings, which is uncountable, so there cannot be a bijection between V and V^*.

Given v\in V, one can define a function f_v:V^*\to \mathbb{F}, as follows: f_v(g):=g(v). It is linear, as f_v(\alpha g+\beta h)=\alpha g(v)+\beta h(v)=\alpha f_v(g)+\beta f_v(h). Here we do not see any dependence on the choice of a basis in V, and we have

The vector space V^{**} of linear functions on V^* is (canonically) isomorphic to V, via the mapping v\mapsto f_v.

Indeed, we see immediately that f_{\alpha v+\beta w}=\alpha f_v+\beta f_w, and so we need only to check that this mapping is bijective. Let \{e_i\} be a basis in V and \{\epsilon_i\} its dual basis in V^*. Then f_{e_i}(\epsilon_j)=1 if i=j and 0 otherwise. Thus \{ f_{e_i}\} is the basis of V^{**}, which is dual to the basis \{\epsilon_i\} of V^*, and the mapping v\mapsto f_v sends the vector with coordinates v_i to the vector with the same coordinates in
the basis \{ f_{e_i}\} of V^{**}. Hence the latter is bijective.

In view of the latter, we can identify V with V^{**}, and write v(g) instead f_v(g). The set of g\in V^* such that v(g)=0 is a subspace, called annihilator of v, of dimension n-1=dim(V)-1. More generally, the following holds.

Let U be a subspace of V, and U^0:=\{g\in V^*\mid g(u)=0\text{ for all } u\in U\}. Then the annihilator U^0 of U is a subspace of V^* of dimension dim(V)-dim(U).

Indeed, we can choose a basis \{e_i\} in V so that \{e_1,\dots,e_{k} is a basis of U, where dim(U)=k. Then we have the dual basis \{\epsilon_i\} of V^*, and U^0 is the subspace with the basis \{e_{k+1},\dots,e_n\}.

In view of this, each U can be obtained as the set of solutions of a system of homogeneous linear equations g(u)=0, for g\in U^0, of rank dim(V)-dim(U).

Dual spaces and annihilators under a basis change
Let X\in GL_n(\mathbb{F}) be a linear transformation of V, and U a subspace of V. Then X(U)=\{ Xu\mid u\in U\} is a subspace. How can one look at g(U^0)? By writing out u=\sum_i u_i e_i in a basis \{e_i\}, for any g=\sum_i g_i\epsilon_i\in U^0 in the dual basis \{\epsilon_i\}, we get equation \sum_i g_i u_i =0. Thus, considering X as a matrix, we get g^T YX u=0, where Y denotes the action of X on V^*. It follows that YX=1_{GL_n}, i.e. Y=X^{-1}. We have, considering that Y acts on V^* by right multiplication, and not by left ones, to take the transpose, too.

X\in GL_n(\mathbb{F}) acts on V^* as (X^{-1})^T.

An example.
Let V=\mathbb{F}^3. We work in the standard basis \{e_1,e_2,e_3\} of V. Then the dual basis of V^* is \{\epsilon_1,\epsilon_2,\epsilon_3\}, so that \epsilon_i((u_1,u_2,u_3)^T)=u_i.
Let G be the group of matrices G=\left\{ \begin{pmatrix} 1&x&y\\ 0&z&u\\ 0&t&w \end{pmatrix} \mid x,y,z,u,t,w\in\mathbb{F} \right\}<GL_3(\mathbb{F}). It fixes, in its left action on V by multiplication, the vector e_1=(100)^T.. Let U be the 1-dimensional subspace of V generated by e_1. Then U^0 is generated by \epsilon_2 and \epsilon_3. The group G preserves U^0, in its action on V^*. As U_0 is 2-dimensional, there should be a nontrivial kernel in this action, and indeed, it consists of the elements of the form \begin{pmatrix} 1&x&y\\ 0&1&0\\ 0&0&1 \end{pmatrix}.

A particularly simple case is \mathbb{F} = \mathbb{Z}_2. Then G is isomorphic to S_4, the symmetric group on 4 letters, as can be seen in its action on the 4 elements of V^* outside U^0. On the other hand, it acts on the 3 nonzero elements of U^0 as S_3.