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Lie Groups/Algebras, Notes 4: Lie Groups

Lie Groups:

Definition 1.  A Lie group is a manifold and group whose binary operation is a C^\omega map between the manifolds G\times G and G.

We will assume our Lie groups are Banach manifolds unless otherwise stated (locally homeomorphic to a Banach space E).  It turns out a Lie group is metrizable, and the metric space is complete.  Lie subgroups are subgroups which are also submanifolds and Lie groups.

Proposition 2.  Let G be an E-Lie group (that is, a Lie group which is an E-manifold).  Then

  1. If E\in\{\mathbb{R},\mathbb{C}\}, then G is locally connected.
  2. If E\notin\{\mathbb{R},\mathbb{C}\}, then \dim(G)=\dim_F(E)=0.
  3. Suppose E is locally compact, then G is locally compact iff \dim(G)<\infty.
  4. If G is generated by a subspace whose topology admits a countable base, then the topology on G admits a countable base.

Now suppose X is a C^\omega-manifold with an analytic associative binary operation with identity.  Then the subset G\subseteq X of invertible elements of X is open in X and is a Lie group.

Definition 3.  A Lie group morphism between Lie groups G and H is a C^\omega group homomorphism f:G\to H.  Aut(G) will denote the group of Lie group automorphisms of G.  A Lie group representation on a Banach space E is a Lie group morphism \rho:G\to Aut(E).  (Note if E is a Banach space over the field F, then Aut(E) is a Lie group which is a manifold over F)

Definition 4.  Let G be a topological group and X be an analytic manifold.  We say that X is a homogeneous space for G if X is a G-space (that is, continuous action) and (g_1g_2)x=g_1(g_2x).

This induces an equivalence relation on X into orbits under the action of G and a corresponding quotient space X/G, called the orbit space of X by G.

Proposition 5.  Let X be Lie group and G be a Lie subgroup.  Then:

  1. The orbit space X/G has a unique analytic manifold structure such that the canonical map \pi:X\to X/G is a submersion.
  2. If G\unlhd X, then X/G is a Lie group and \pi is a Lie group morphism.

In case 1 above, we call the quotient X/G the Lie homogeneous space of X by G, and in case 2 we call the quotient the Lie quotient group.

Let G and H be Lie groups.  Then there is a canonical correspondence \tau between T(G)\times T(H) and T(G\times H).  In particular, suppose we have two tangent vectors u\in T_x(G), v\in T_y(G), and f\in C^\omega(G).  Let us then define


It follows that uv\in T_{xy}(G) for

\begin{array}{lcl} (uv)(fg)&=&\tau(u,v)(fg,0)+\tau(u,v)(0,fg)\\&=&\tau(u,v)(f,0)\cdot g(xy)+f(xy)\tau(u,v)(g,0)+\tau(u,v)(0,f)\cdot g(xy)+f(xy)\tau(u,v)(0,g)\\&=&g(xy)\left(\tau(u,v)(f,0)+\tau(u,v)(0,f)\right)+f(xy)\left(\tau(u,v)(g,0)+\tau(u,v)(0,g)\right)\\&=&(uv)(f)g(xy)+f(xy)(uv)(g).\end{array}

Thus we have a product on the tangent bundle of G (note that G being a group makes xy an arbitrary element of G).  Hence T(G) is a Lie group.

The Lie Algebra of a Lie Group:

Recall a vector field on a manifold M is section of the vector fiber bundle T(M) with M as its base.  Hence it is a map X:M\to T(M) satisfying \pi\circ X=1 where \pi is the projection of the vector fiber bundle to its base.  Now suppose X and Y are two vector fields on M, then we can define the vector fields


(XY)(p)(f)=X(p)[Y\circ f]-Y(p)[X\circ f]

where (Y\circ f)(x)=Y(x)(f).  This turns V(M), the set of vector fields of M, into a Lie algebra.

Now let G be a Lie group and \lambda_g:G\to G be left multiplication by g\in G.  These are certainly diffeomorphisms of G

Definition 6.  A vector field X is left-invariant if


for all g\in G where (\lambda_g)_*(X)(x)=X(gx).

Proposition 7.  The set L(G) of left-invariant vector fields of G is a Lie subalgebra of V(G).

Proof.  Let X and Y be left-invariant and g\in G.  Then



\displaystyle \begin{array}{lcl}(\lambda_g)_*(XY)(x)(f)&=&(\lambda_g)_*\left(X(x)[Y\circ f]-Y(x)[X\circ f]\right)\\&=&X(gx)[Y\circ f]-Y(gx)[X\circ f]\\&=&(XY)(gx)(f)\end{array}.

The Lie subalgebra L(G) of V(G) is called the Lie algebra of G.  Note an obvious consequence is that if v\in T_1(G), X\in L(G), and X(1)=v.  Then


Hence it follows that left-invariant vector fields of G are completely determined by where they send the identity element.  In this sense, we obtain the equivalence of the vector spaces L(G) and T_1(G).

Lie Group of a Lie Algebra:

The standard convention is to discuss an “inverse” map from the Lie algebra of a Lie group back to the Lie group.  Let X\in L(G).  Then this left-invariant vector field corresponds to a tangent vector v\in T_1(G).  Let \gamma_X:\mathbb{R}\to G be a path defined by




Note these uniquely define \gamma_X, called a one-parameter subgroup of G.  We in turn define the map \exp:L(G)\to G by


or more generally


Hence \exp(0)=1. It follows that L(\exp(L(G)))=L(G).

Suppose G=GL_n(\mathbb{R}).  Then L(G)=M_n(\mathbb{R}) and


[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.

[2]  Robert Milson, Thomas Foregger, Mike Fikes. “Lie group” (version 12). Freely available at;from=objects;id=1112


Lie Groups/Algebras, Notes 3: Bialgebras and Free Lie Algebras

Recall the construction of an associative algebra A over a field K.  We have a module homomorphism \nabla:A\otimes A\to A such that

\nabla\circ (1\otimes\nabla)=\nabla\circ(\nabla\otimes 1).

If the algebra is unital, then we also a canonical map \eta:K\to A with \eta(1) defined as the identity of A and corresponding assumption that

\nabla\circ(1\otimes\eta)=\nabla\circ(\eta\otimes 1)=1

on A (with canonical identifications A\otimes K=K\otimes A=A).

We can dually define a Kcoalgebra as a K-module C together with a comultiplication map \Delta:C\to C\otimes C such that

(1\otimes\Delta)\circ\Delta=\Delta\circ(\Delta\otimes 1).

We also call it unital if there is a map \varepsilon:C\to K such that

(1\otimes\varepsilon)\circ\Delta=(\varepsilon\otimes 1)\circ\Delta=1

on C (with the canonical identification C\otimes K=K\otimes C=C).

Definition 1.  A Kbialgebra is a unital K-algebra and unital K-coalgebra such that

  1. \Delta\circ\nabla=(\nabla\otimes\nabla)\circ(1\otimes\tau\otimes 1)\circ(\Delta\otimes\Delta) (\tau(x\otimes y)=y\otimes x),
  2. \varepsilon\circ\nabla=\varepsilon\otimes\varepsilon,
  3. \Delta\circ\eta=\eta\otimes\eta, and
  4. \varepsilon\circ\eta=1.

Definition 2.  An element x of a coalgebra is uprimitive if \Delta(x)=x\otimes u+u\otimes x.  Elements are primitive if they are 1-primitive.

If \sigma:\mathfrak{g}\to U(\mathfrak{g}) is the embedding into the enveloping algebra, then the map \Delta(\sigma(x))=\sigma(x)\otimes 1+1\otimes\sigma(x) gives U(\mathfrak{g}) a K-bialgebra structure.

Definition 3.  Let X be a set and A_K(X) be the free algebra over X.  We define the free Lie Kalgebra over X as the quotient A_K(X)/I where I is the ideal generated by elements of the form xy+yx and x(yz)+z(xy)+y(zx).  We will denote this Lie algebra by L(X) and its product by [x,y].

The enveloping algebra U(L(X)) of the free Lie algebra L(X) ends up being the free associative algebra over X.

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.

Lie Groups/Algebras, Notes 2: Representations and Types

Recall a representation of an algebra A on an R-module M is an algebra morphism \rho:A\to End_R(M).  When we have a Lie algebra \mathfrak{g}, a representation of it on an R-module M is an algebra morphism

\rho:\mathfrak{g}\to L(End_R(M)).

Hence in particular we have \rho(xy)(m)=\left(\rho(x)\rho(y)-\rho(y)\rho(x)\right)(m).  We correspondingly call M a \mathfrak{g}module.  A representation is faithful if it is injective.  It is simple/irreducible if its \mathfrak{g}-module is simple.

Let \rho_1 and \rho_2 be representations of Lie algebras \mathfrak{g}_1 and \mathfrak{g}_2 on M_1 and M_2 respectively.  Then \rho_1\otimes\rho_2 is a representation of \mathfrak{g}_1\times\mathfrak{g}_2 on M_1\otimes M_2 defined by

\begin{array}{lcl}(\rho_1\otimes\rho_2)(x_1,x_2)(m_1\otimes m_2)&=&(\sigma_1(x_1)\otimes 1+1\otimes\sigma_2(x_2))(m_1\otimes m_2)\\&=&\sigma_1(x_1)(m_1)\otimes m_2+m_1\otimes\sigma_2(x_2)(m_2)\end{array}

where \sigma_1,\sigma_2 are the canonical inclusions of \mathfrak{g}_1 in U(\mathfrak{g}_1) and \mathfrak{g}_2 in U(\mathfrak{g}_2).  Hence M_1\otimes M_2 is both a \mathfrak{g}_1\times\mathfrak{g}_2-module and a U(\mathfrak{g}_1)\otimes U(\mathfrak{g}_2)-module.

Definition 1\mathfrak{g} is solvable if its derived series D^n\mathfrak{g}=0 for some n.  It is nilpotent if its lower central series C_n\mathfrak{g}=0 for some n.  The radical of \mathfrak{g}, \sqrt{\mathfrak{g}}, is the largest solvable ideal in \mathfrak{g}.  We say \mathfrak{g} is semisimple if it has no nonzero abelian ideals.

Hence it follows that the center Z(\mathfrak{g}) of a nilpotent Lie algebra is nontrivial.  Also note that a nilpotent Lie algebra is solvable.  \mathfrak{r}=\sqrt{\mathfrak{g}} is the smallest ideal such that \mathfrak{g}/\mathfrak{r} has radical 0.

Theorem 2 (Engel).  Let V be a vector space and \mathfrak{g} be a finite dimensional subalgebra of \mathfrak{gl}(V)..  If V\neq 0, then there exists a u\neq 0 with u\in\mathfrak{g} such that xu=0 for all x\in\mathfrak{g}.

It follows that a Lie algebra \mathfrak{g} is nilpotent iff \mbox{adj}\,x is nilpotent for all x\in\mathfrak{g}.

Note that B(x,y)=\mbox{tr}(\mbox{adj}(x)\mbox{adj}(y)) defines a symmetric bilinear form on \mathfrak{g}, called the Killing form of \mathfrak{g}.  The Killing form also satisfies B(xy,z)=B(x,yz).

Proposition 3.  \mathfrak{g} is solvable iff B(D\mathfrak{g},\mathfrak{g})=0.

Proposition 4.  \mathfrak{g} is semisimple iff \mathfrak{g}=0 iff B is nondegenerate.

Theorem 5 (Weyl).  Every finite dimensional representation of a semisimple Lie algebra is completely reducible (i.e. the \mathfrak{g}-module is semisimple).

A Lie algebra is simple if its only ideals are \mathfrak{g} and 0 and noncommutative.

Proposition 6.  \mathfrak{g} is semisimple iff it is a product of simple algebras.

A Lie algebra is reductive if its adjoint representation is semisimple.  That is, it is reductive if it can be written \mathfrak{g}=\mathfrak{s}\oplus\mathfrak{a} where \mathfrak{s} is semisimple and \mathfrak{a} is abelian (since here we want \mathfrak{g} as a module to be semisimple–so that as an algebra, it has an an abelian term).  This also turns out to be equivalent to D\mathfrak{g} being semisimple.

Definition 7.  The nilpotent radical of \mathfrak{g} is the intersection of the kernels of all finite dimensional simple representations of \mathfrak{g}.

For an element x in the nilpotent radical, it follows that \mbox{adj}(x) is nilpotent.  Now for any nilpotent endomorphism u of an algebra, we have that

\displaystyle e^u=\sum_{n=0}^\infty\frac{u^n}{n!}

has a finite number of nonzero terms, and is hence well-defined.  If we can write \mathfrak{g}=\sqrt{\mathfrak{g}}\ltimes_{e^{u\cdot}}\mathfrak{s} with \mathfrak{s} semisimple and u nilpotent, then we will call \mathfrak{s} a Levi subalgebra of \mathfrak{g}.

Theorem 8 (Levi-Malcev).  Every Lie algebra has a Levi subalgebra (that is, it can be written \mathfrak{g}=\sqrt{\mathfrak{g}}\ltimes\mathfrak{s} as above).  Moreover, every other Levi subalgebra \mathfrak{s}' is the image of \mathfrak{s} under \exp(\mbox{adj}(x)) for some x in the nilpotent radical of \mathfrak{g}.

It turns out that a subalgebra of \mathfrak{g} is a Levi subalgebra iff it is a maximal semisimple subalgebra.

Theorem 9 (Ado).  Let \mathfrak{n} be the largest nilpotent ideal in \mathfrak{g}.  Then \mathfrak{g} admits a finite dimensional faithful representation \rho such that every element of \rho(\mathfrak{n}) is nilpotent.

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.

Lie Groups/Algebras, Notes 1: Lie Algebras

Definition 1.  A Lie algebra \mathfrak{g} is an R-algebra such that xy+yx=0 and x(yz)+z(xy)+y(zx)=0.

Hence x^2=-x^2\Rightarrow 2x^2=0\Rightarrow x^2=0 provided R is not a ring of characteristic 2.  Also, if A is an R-algebra, then we can define a Lie algebra L(A) whose product is defined by


Hence in particular, L(\mathfrak{g}) simply doubles the product of \mathfrak{g}.  Recall that a derivation d on an algebra satisfies




For x\in\mathfrak{g}, let us define the adjoint of x as \mbox{adj}_x\in\mbox{End}\mathfrak{g} defined by


Proposition 2.  For x\in\mathfrak{g}, \mbox{adj}_x is a derivation on \mathfrak{g}.

Proof.  By tensorial properties of algebras we have




 Proposition 3.  Let \mathfrak{g} be a commutative Lie algebra over a ring R with \mbox{char}(R)\neq 2.  Then xy=0.  If xy=0, then \mathfrak{g} is commutative, regardless of the ring.

Proof.  If \mathfrak{g} is commutative, then xy=yx=-xy\Rightarrow 2xy=0\Rightarrow xy=0.  The second claim is trivial.

Let I be a left ideal in \mathfrak{g}, x\in I, and y\in\mathfrak{g}.  Then yx=-xy\in I and hence xy\in I.  So I is also a right ideal.  Hence we simply consider ideals of Lie algebras without reference to leftness or rightness.

Definition 4.  Let I be a submodule of \mathfrak{g} such that dI\leq I for all derivations d on \mathfrak{g}, then I is called a characteristic ideal of \mathfrak{g}.

It’s easy to see that a characteristic ideal is actually an ideal in \mathfrak{g} since we have

\mbox{adj}_xI=xI\leq I

for all x\in\mathfrak{g}.

Proposition 5.  Let \mathfrak{a} be an ideal of \mathfrak{g} and \mathfrak{b} be a characteristic ideal of \mathfrak{a}, then \mathfrak{b} is an ideal of \mathfrak{g}.

Proof.  If d is a derivation on \mathfrak{a}, then d\mathfrak{b}\leq\mathfrak{b} since \mathfrak{b} is a characteristic ideal.  But multiplication by x\in\mathfrak{g} is a derivation on \mathfrak{a}.  Hence bx\in\mathfrak{b} for b\in\mathfrak{b}.

For two ideals \mathfrak{a},\mathfrak{b}\subseteq R, we have that \mathfrak{a}+\mathfrak{b} and \mathfrak{a}\cap\mathfrak{b} are ideals in R and that \mathfrak{a}\cup\mathfrak{b}\subseteq\mathfrak{a}+\mathfrak{b} and \mathfrak{a}\mathfrak{b}\subseteq\mathfrak{a}\cap\mathfrak{b}.  In the case of an R-algebra A, we will let \mathfrak{a}\mathfrak{b} denote the submodule generated by elements ab.  This is in fact an ideal in a Lie algebra for if d is a derivation, then


Hence it is stable under adjoint action.

Definition 6.  The ideal D^0\mathfrak{g}=\mathfrak{g}\mathfrak{g} is called the derived ideal of \mathfrak{g}.  Inductively we can define


The sequence \{D^n\mathfrak{g}\} is called the derived series of \mathfrak{g}.  Let us define C_1\mathfrak{g}=\mathfrak{g} and


The sequence \{C_n\mathfrak{g}\} is called the lower central series of \mathfrak{g}.

Definition 7.  Let \mathfrak{g} be a Lie algebra, T(\mathfrak{g}) be its tensor algebra (with \mathfrak{g} interpreted as a module), and I be the ideal generated by elements of the form x\otimes y-y\otimes x-xy.  Then L^{-1}(\mathfrak{g})=T/I is called the enveloping algebra of \mathfrak{g}.

Theorem 8.  The enveloping algebra L^{-1}(\mathfrak{g}) satisfies the universal property that if f:\mathfrak{g}\to L(A) is an algebra homomorphism for some associative unital algebra A and \iota:\mathfrak{g}\to L(L^{-1}(\mathfrak{g})) is the canonical inclusion, then there exists a unique morphism g:L^{-1}(\mathfrak{g})\to A and \ell(g):L(L^{-1}(\mathfrak{g}))\to L(A) such that f=\ell(g)\circ\iota.

Proposition 9.  If \mathfrak{g}_1 and \mathfrak{g}_2 are Lie algebras, then L^{-1}(\mathfrak{g}_1\times\mathfrak{g}_2)=L^{-1}(\mathfrak{g}_1)\otimes L^{-1}(\mathfrak{g}_2).

Let us define U_n(\mathfrak{g})=T^n(\mathfrak{g})/I with I defined as above, G_n=U_n/U_{n-1}, and G=\oplus G_n.  It follows that G is a filtered algebra, which we will call the associated filtered algebra to L^{-1}(\mathfrak{g}).

Theorem 10 (Poincare-Birkhoff-Witt)  Let G be the associated filtered algebra to L^{-1}(\mathfrak{g}) and S(\mathfrak{g}) be the symmetric algebra of \mathfrak{g} as a module.  Then if \mathfrak{g} is free, S(\mathfrak{g})=G.

It follows that if \mathfrak{g} is free, then the canonical homomorphism f:\mathfrak{g}\to L^{=1}(\mathfrak{g}) is injective.

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.

Commutative Algebra, Notes 6: Valuations

Definition 1.  Let F be a field.  An absolute value on F is a map |\cdot|:F\to\mathbb{R} such that for all x,y\in F

  1. |x|\geq 0 and |x|=0 iff |x|=0.
  2. |xy|=|x||y|.
  3. |x+y|\leq|x|+|y|.

We will call the pair (F,|\cdot|) an absolute field.  The absolute value that sends all nonzero elements to 1 is called the trivial absolute value.  One can also verify that the function d(x,y)=|x-y| defines a metric on F .  Two absolute values are dependent if their induced topologies are the same, and independent otherwise.  Note that

|1|=|1^2|=|1|^2 and |1|=|(-1)^2|=|-1|^2.

Hence |1|=1, and thus |-1|=\sqrt{1}=1 since |x|\geq 0.  Similarly


and 1=|1|=|xx^{-1}|=|x||x^{-1}|\Rightarrow |x^{-1}|=|x|^{-1}.

Proposition 2.  Let |\cdot|_1 and |\cdot|_2 be two nontrivial absolute values on F .  Then |\cdot|_1 and |\cdot|_2 are dependent iff |x|_1<1\Rightarrow |x|_2<1.  If they are dependent, then there exist some \lambda>0 such that |\cdot|_1=|\cdot|_2^\lambda.

Theorem 3.  (Approximation Theorem) (Artin-Whaples).  Let F be a field and |\cdot|_1,...,|\cdot|_n be nontrivial pairwise independent absolute values on F .  Then there exist elements x_1,...,x_n\in F such that for any \varepsilon>0, there exists an x\in F such that


for all 1\leq i\leq n.

An absolute field is complete if all of its Cauchy sequences converge to a point in the field.  One can show that each absolute field has a unique completion field up to isometry.

Definition 4.  An abelian group G is an ordered group if it has a partial ordering such that x\leq y\Rightarrow xg\leq yg for all x,y,g\in G.

Proposition 5.  (G,\leq) is an ordered group iff G=\{1\}\sqcup S\sqcup S^{-1} for a multiplicative subset S\subset G where x<1<y for all x\in S and y\in S^{-1}.

For an ordered group G, we will hereafter use G to denote G[0] where 0g=0 and 0<g for all g\in G.

Definition 6.  Let F be a field and G be an ordered group.  A valuation on F is a map |\cdot|:F\to G such that

  1. |x|=0 iff x=0.
  2. |xy|=|x||y|.
  3. |x+y|\leq\max\{|x|,|y|\}.

We will call the triple (F,G,|\cdot|) a valuation field.  The image of |\cdot| on F is an ordered subgroup of G.  Two valuations are equivalent if there is an isomorphism \lambda:F\to G that respects order and value.  One can also verify that |\pm x|=|x| and that if |x|<|y|, then |x+y|=|y|.

Definition 7.  A subring R of a field F is a valuation ring if for all x\in F, x\in R or x^{-1}\in R.

Proposition 8.  If R\leq F is a valuation ring, then it induces a valuation field.

Proof.  Note that R is a local ring, for, if x,y are not units in R, then if x/y\in R,  we have

1+x/y=(x+y)/y\in R.

So if x+y is a unit in R, then 1/y\in R, which is a contradiction.  Hence x+y is not a unit.  Furthermore if x is not a unit, then for all r\in R we have that rx is not a unit.  So the nonunits form an ideal in R, which is necessarily maximal (and uniquely so).  So R is a local ring.  Let us denote this ideal by \mathfrak{m}.  Thus we can write

F^*=\mathfrak{m}^*\sqcup U\sqcup\mathfrak{m}^{*^{-1}}

where U is the group of units of R.  We now give F a valuation.  We will define our group G as the quotient group F^*/U.  In turn, we define |x|=xU with |1|=U:=1.  When then define |0|=0, |x|<1 iff x\in\mathfrak{m}^*, and |x|>1 iff x\in\mathfrak{m}^{*^{-1}}.  Observe that




Hence (F,F^*/U,|\cdot|) is a valuation field.

We thus hereafter refer to a valuation ring as the restriction of the induced valuation field to the ring.

Proposition 9.  Let F\leq K be fields where F is a valuation field.  Then there exists an extension making K a valuation field.

Proof Idea.  Let R=\{x\in F:|x|\leq 1\} and U=\{x\in F:|x|=1\}.  We will call R the valuation ring of F .  One first takes the morphism \varphi:R\to R/(R-U) and extends it to a valuation ring D in K.  One can construct an order preserving monomorphism

\psi:F^*/U\to K^*/U'

where U' is the maximal ideal in D.  We then define |x|_K as before, which is seen to agree with |x|_F.

Proposition 10.  Let F\leq K be fields and [K:F]=n.  Let F be a G-valued valuation field and G' be the extension group to the induced valuation on K.  Then [G':G]\leq n.

We will call [G':G] the ramification index of F in K.

Definition 11.  A valuation is discrete if its codomain is a cyclic group.

It turns out that if F is a valuation field and \mathfrak{m} is the maximal ideal of its valuation ring, then there exists an element \pi\in\mathfrak{m} such that |\pi| is a generator of G.  Such an element is called a local parameter of |\cdot|.  One can also show that \mathfrak{m} is a principal ideal generated by \pi.  Moreover every element x\in F can be written as


for some unit u\in R integer r.  r is called the order of x at |\cdot|.  If r>0, we say x has a zero of order r, and if r<0, then we say x has a pole of order r.

Proposition 12.  Let F\leq K be fields where K is a finite extension of F .  Further suppose that F is a complete discrete valuation field (i.e. induced metric space is complete) and that R,R' and \mathfrak{m},\mathfrak{m}' are the corresponding valuation rings and maximal ideals after extending the valuation.  Then


 [1] Lang, Serge.  Algebra.  Revised Third Edition.  Springer-Verlag.  2000.