This is a graduate course in rings and algebras, covering parts II (Ring Theory) and III (Modules and Vector Spaces) of Dummit and Foote.
This chapter covers the basics of ring theory: rings, ideals, homomorphisms, fractions, and many standard examples.
There are plenty of definitions, and one way to remember them is by using examples that you already know. So instead of remembering the multipart definition of rings, remember that \( \mathbf{Z}/n\mathbf{Z} \) is a commutative ring, \( \mathbf{Q} \) is a field, and the ring of polynomials \( \mathbf{R}[x] \) is an integral domain. Further, \( \mathbf{H} \) is a division ring and the matrix ring \( M_n(\mathbf{C}) \) is not.
There are some little quirks in the definitions that you need to be aware of: for example, zero is not a zero-divisor.
A very important example is introduced: the ring of algebraic integers in a number field. This is not formally defined, but the important examples \( \mathbb{Q}[\sqrt{D}] \) (a number field) and \( \mathbb{Z}[\sqrt{D}] \) (a ring) are given.
I'm going to assume you know what polynomial and matrix rings are, because you are forced to teach calculus and linear algebra. A group ring \( RG \) is defined to be the set of linear combinations \( a_1g_1 + a_2 + \cdots + a_ng_n \), where the \(a_i\) live in a ring \( R \) and the \(g_i \) in a group \( G \). These are a little bit stranger: there will generally be zero-divisors.
A model for a ring homomorphism is the reduction mod \(n \) which maps \(\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z} \). The kernel of this homomorphism is the ideal \(n\mathbb{Z}\). There are four standard properties of homomorphism and ideals, which you would guess that they have based on the analogy with groups. Below, \( \phi \colon R \rightarrow S \) is a ring homomorphism, \(A \subset R\) is a subring, and \( I , J \) are ideals.
An ideal is an analog of a normal subgroup, that is, one can take the quotient ring \( R/I \) of a ring \( R \) by an ideal \( I \subset R \). An ideal is also an analog of a subspace (or: it is a submodule), that is, one can take the ideal generated by a set of elements \( \{ a_1, \ldots, a_n \} \) in \( R \), and this is (at least in the commutative case) the set of all linear combinations \( r_1 a_1 + \cdots + r_n a_n \), with coefficients \( r_i \in R \). An ideal could be generated by a single element (principal ideal), a finite number of elements, or an infinite collection of elements.
You should keep in mind a few examples of ideals and quotient rings. The most obvious one is \( \mathbb{Z} / n \mathbb{Z} \). Another is \( F[x] / ( p ) \), where \( p \in F[x] \) is a polynomial. The ideal of continuous functions on the real line that vanish at a point (say zero) is not finitely generated.
There is a connection between properties of an ideal and properties of the corresponding quotient ring. Two important examples of this: an ideal is maximal (with respect to containment) iff the quotient ring is a field. An ideal is prime iff the quotient ring is an integral domain. The definition of prime is a generalization of the concept of prime number. Maximal ideals are prime, but not the other way around: the ideal \( (y) \subset F[x,y] \) is prime but not maximal.
Zorn's Lemma can be used to prove that every ideal is contained in a maximal ideal.
The construction of \( \mathbb{Q} \) from \( \mathbb{Z} \) is to allow fractions \( \frac{a}{b} \) where \( a \in \mathbb{Z} \) and \( b \neq 0 \), declaring some of these fractions to be equivalent to others in a familiar way. This works with a ring \( R \) and a set of denominators \( D \) if the set of denominators is closed under multiplication, does not contain zero, and does not have zero-divisors. This is sometimes called localization, especially in the context of algebraic geometry. The term localization is used because \( R \) a ring of functions defined on some geometric object and one wants to study this object locally near a point \( p \), then one takes \( D \) to be all functions that are not zero at \( p \).
A simple version of this theorem, which might be easier to remember: if we know an integer mod 4 and mod 3, then we know the integer mod 12. A fancier way to say this is that \( \mathbb{Z}/12 \mathbb{Z} \cong \mathbb{Z}/4 \mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z} \). Of course there is a much more general version.
A norm is a measure of size on the elements of a ring; this size is required to be a positive integer (except for \( N(0) \)). A Euclidean Domain is one with a norm which works in concert with division: if \( a, b \in R \), we can divide \(a\) by \(b\), which results in a quotient \(q\) and remainder \(r\). This is trivially possible with \( q = 0 \) and \( r = a\), so the key extra condition which defines a Euclidean Domain is that we can do this with \( N(r) < N(b) \).
The two main examples of Euclidean Domains are \( \mathbb{Z} \) and \( F[x] \). These are also PIDs, by the main result of this chapter. The idea of the proof is the extreme principle: consider a nonzero element of minimal norm in the ideal.
A PID is a commutative ring in which every ideal is principal, i.e. generated by a single element. The rings \( \mathbb{Z}[x] \) and \( F[x,y] \) are not PIDs. An example of a PID which is not a Euclidean Domain is \( \mathbb{Z}[(1 + \sqrt{-19})/2] \). This is shown (in 8.1) by observing that Euclidean Rings have universal side divisors, which do not exist in \( \mathbb{Z}[(1 + \sqrt{-19})/2] \).
A universal side divisor is an non-zero, non-unit element which divides any element of the ring with remainder either a unit or zero. For example, in \( \mathbb{Z} \), we can write any number \(n \) as \( q 2 + r \), where \( r = 0 \text{ or } 1 \), so \( 2 \in \mathbb{Z} \) is a universal side divisor. Generally any non-unit, non-zero element of minimal norm in a Euclidean Ring is a universal side divisor.
To see that such things do not exist in \( \mathbb{Z}[(1 + \sqrt{-19})/2] \), use the norm to show that the only units are \( \pm 1 \). Then a universal side divisor \( u \) would have to divide \(2\) up to a unit, so it would have to divide \( \pm 2 \text{ or } \pm 3 \). Again using the norm, one can show that the only such non-unit elements are \( \pm 2 \text{ or } \pm 3 \). But these do not divide \( [(1 + \sqrt{-19})/2] \pm 1 \).
An integral domain is a PID iff it has a Dedekind-Hasse norm. The definition requires that for any \(a, b \in R \), either \( a \in (b) \) or there is an element \( sa-tb \in (a,b) \) with norm less than \(N(b)\). This generalizes the division condition in the definition of Euclidean by allowing \( s \neq 1 \).
The idea of the proof that a ring with a DH norm is a PID is again to use the extreme principle. The proof of the converse proceeds by using the fact that a PID is a UFD. The function \(N(a) = 2^n \), where \(n\) is the number of irreducibles in the factorization of \( a \), is a DH norm.
This type of ring is a way of generalizing or axiomatizing unique factorization into primes in \( \mathbb{Z} \). As with \( \mathbb{Z} \), the factorization is only defined up to order and units. The key definition is irreducible. An element \( p \in R \) is irreducible if any factorization \( p = a b \) has either \(a \) or \( b \) a unit.
This is different from the definition of a prime element: one for which \( (p) \) is a prime ideal. Prime elements (of integral domains) are always irreducible, but not the other way around. For example, in \( \mathbb{Z}[\sqrt{-5}] \), \( 3 \) is irreducible, but \( (1 + \sqrt{-5})(1-\sqrt{-5}) = 6 \in (3) \), so \( (3) \) is not prime.
The proof that a PID is a UFD requires showing that a factorization is somehow finite. This is not obvious. For example in the ring of continuous functions on \( [0,1] \) we have \( x = x^{\frac{1}{n}} \cdots x^{\frac{1}{n}} \) for any \( n \). This can't happen in a PID, because it would give rise to an infinite ascending chain of ideals, and the union of these ideals would be generated by a single element. Once it is shown that factorizations are finite, the uniqueness is analogous to the proof of unique factorization in \( \mathbb{Z} \): cancel the factors, one by one.
This chapter is a collection of results on polynomial rings, which are studied more deeply later in the book. A major theme is induction, whether on the degree of a polynomial or the number of variables in the polynomial ring. This allows us to show that many properties of \(R \) are inherited by \( R[x_1, \ldots, x_n] \).
If \( R \) is an integral domain, it follows that \(R[x]\) is also, and that the units of \( R[x] \) are the units of \(R\). As indicated, this extends to \( R[x_1, \ldots, x_n] \).
If \( I \subset R \) is an ideal, we can form the rings \( (R/I)[x] \) and \( R[x]/ I[x] \); they are isomorphic. From the previous paragraph it follows that \( I[x] \subset R[x] \) is prime iff \( I \) is prime. The example \( 2[x] \subset \mathbb{Z}[x] \) shows that this is not true with prime replaced by maximal.
There are two main results in this section: Gauss's Lemma, which says that if \( F \) is the field of fractions of a UFD \( R \), then a polynomial \( p(x) \in R[x] \) factors in \( R[x] \) iff it factors in \( F[x] \).
This result requires unique factorization. The textbook gives the follwing counterexample (to Gauss's Lemma for integral domains) \( R = \mathbb{Z}[2i] \) , \( F = \mathbb{Q}[i] \), \( p = x^2 + 1 \). In \(R[x] \), \( p \) is irreducible. But in \(F[x]\), \( p = (x-i) (x+i) \).
The proof of the lemma, after taking the natural step of clearing denominators, uses the indentification of irreducible and prime in a UFD.
The other result is that \(R[x]\) is a UFD iff \(R \) is.
One of the most common ways to show that a polynomial is irreducible is the Eisenstein Criterion. If the coefficients have certain divisibility properties, these divisibility properties cannot be reproduced by a product. Here are the divisibility properties, for a polynomial \( a_n x^n + \cdots a_1 x + a_0 \in R[x] \), where \( P \subset R \) is a prime in an integral domain:
The canonical example of applying the criterion is the proof that the cyclotomic polynomial \( x^{p-1} + x^{p-2} + \cdots + x + 1 \) is irreducible ( \( p \) prime ). Make the change of variable \( y = x - 1 \), check for factors of \( p \) in binomial coefficients, and apply the criterion.
Warning: there are lots of tricks for showing that specific polynomials are irreducible. For example, a polynomial of degree 3 that doesn't have a root must be irreducible. However, all of these things are merely tricks. There are lots of polynomials which are irreducible, but whose irreducibility does not follow from the tricks.
The structure of the multiplicative groups of \( \mathbb{Z}/n\mathbb{Z} \) and \( F[x] / (f) \) is discussed. By the Chinese Remainder Theorem, you can break it up into a product corresponding to prime powers, and then it turns out that \( \mathbb{Z}/2^n \mathbb{Z} \cong C_2 \times C_{2^{n-2}} \) for \( n > 2\) and that \( \mathbb{Z}/p^n \mathbb{Z} \) is cyclic of order \(p^{n-1}(p-1)\) for \(p > 2\).
This is a long section, and the ideas are less likely to make it to an undergraduate algebra course, so the summary will be more detailed here.
A Noetherian Ring is one in which there are no infinite ascending chains of ideals. Equivalently, every ideal is finitely generated. This class of rings includes polynomial rings in \( n \) variables over a field, because of the Hilbert Basis Theorem, which says that \( R[x]\) is Noetherian iff \(R \) is.
The ascending chain condition and descending chain condition might seem to be symmetric, but they are not. This is because in a ring we can multiply but not divide or take roots. More specifically, if \( x \in R \), we can instantly write down something that looks like an infinite descending chain: \( R \supset (x) \supset (x^2) \supset (x^3) \supset \cdots \). We cannot in a general ring write down the analogous ascending chain \( (x) \subset (x^{\frac{1}{2}}) \subset (x^{\frac{1}{3}}) \subset \cdots \subset R \).
The proof of the Hilbert basis theorem takes an ideal \( I \subset R[x]\) and looks at ideals (in \(R\)) of coefficients of leading terms of polynomials in \( I \). This motivates an extensive discussion of leading terms and an introduction to the theory of Grobner Bases.
The concept of leading term for a polynomial in many variables is not defined until we introduce a monomial order. The most obvious ordering on monomials is the lexicographic order, which chooses some order on the variables and then orders monomials as in a dictionary. Having done this we can define leading term (LT) and for an ideal \( I \), the ideal \( \text{LT}(I) \) of leading terms of elements of \( I \). The ideal of leading terms is not always obvious from looking at the polynomials, and may depend on the monomial order.
The definition of a Grobner Basis of an ideal \( I \) is a set of generators \( ( g_1, \ldots, g_m ) \) for the ideal such that \( \text{LT}(I) = ( \text{LT}(g_1), \ldots, \text{LT}(g_m) ) \). The key property is that every element \( f \) of the polynomial ring can be expressed uniquely in the form \( f = q_1 g_1 + \cdots + q_m g_m + r \), in a variant of the division algorithm.
This division algorithm doesn't always have great properties, but using it with Grobner bases gives it somewhat better properties. In particular, we get the uniqueness, and the order we put on the dividing polynomials does not affect the result.
Grobner bases always exist, and many things can be done with them in an algorithmic fashion. For example, it is possible to write a program to check whether two ideals are the same or not (Corollary 28). Some computer algebra systems enable computations with these bases, for example Macaulay2 and Singular.
A minimal Grobner basis is one in which the polynomials are normalized so that the coefficients of the leading terms are 1 and has no redundant elements. A reduced Grobner basis is a minimal one in which no term of \( g_i \) is divisible by \( \text{LT}(g_j) \) for \( j \neq i \). Reduced Grobner bases are unique, which partly accounts for the power of the technique.
Elimination ideal [more]
A module \( M \) over a ring \( R \) is much like a vector space over \( R \), except that the definition of vector space requires a field (or at least a division ring). In particular, if \( F \) is a field, then modules over \( F \) and vector spaces over \( F \) are the same thing. There is also some analogy to the action of a group \( G \) on a set \( X \): the ring is like \( G \) and the module is like \( X \).
The definition says that we can add elements of the module and multiply these module elements by elements of the ring. Just as with vector spaces there are a bunch of axioms for this. The first example is that \( \mathbb{Z} \)-modules are abelian groups. This shows that things are different from the vector space case. Finite dimensional vector spaces over \( F \) are all isomorphic to \( F^n \), and we can form the analogous \( \mathbb{Z} \)-modules \( \mathbb{Z}^n \). But there are also modules like \( \mathbb{Z}/n\mathbb{Z} \) which have no analog in the context of vector spaces.
Another important example is that if we have a vector space \( V \) over \( F \) and a linear transformation \( T \colon V \rightarrow V \) then we can form a \(F[x]\) module by defining the multiplication \( p(x) \cdot v \) to be \( p(T)(v) \). This allows us to unify the Jordan Canonical Form and the Fundamental Theorem of Finite Abelian Groups: both are special cases of the classification of modules over a PID.
Modules are like abelian groups in the sense that given a module and a submodule we can form a quotient: no extra normality condition is required. The four isomorphism theorems which you would expect from the analogy with groups and rings are true: [fix]
Many of the concepts in vector spaces carry over in some form to modules. In particular, there is a concept of direct sum via ordered pairs, just as for vector spaces. The concept of spanning set has an analog: generating set. If the elements of the generating set satisfy the analog of linear independence (no linear relation with coefficients in the ring), then the module is said to be free on the generating set. Generating sets of free modules have a universal property which is shared by bases of vector spaces: you can define a homomorphism by giving its values on the generating set.
If we had a vector space \(V \) with basis \( \{ e_1, \ldots, e_n \} \) and a vector space \( W \) with basis \( \{ f_1, \ldots, f_m \} \), we could form an \( mn \)-dimensional vector space with basis elements the symbols \( e_i \otimes f_j \). The tensor product is a generalization of this bilinear construction to modules. In what follows I will suppress details of the construction, not worrying about the ring over which we have a module structure, nor whether it is on the left, right, or both. An important fact to remember: the typical element of the tensor product \( V \otimes W \) is not \( v \otimes w \) but a linear combination \( \sum c_{ij} e_i \otimes f_j \).
Many properties that you would expect by analogy with the vector space case do in fact hold. For example, \( \mathbb{Z}^2 \otimes \mathbb{Z}^3 \cong \mathbb{Z}^6 \). However, the axioms for bilinearity have some not-entirely-intuitive consequences, like \( \mathbb{Z}/3\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z} \cong 0 \).
Instead of worrying about the construction and the proof of the properties of tensor product, I suggest instead memorizing a series of facts, which will give you the right intuition. One fact that is very intuitive is that \( \otimes \) acts like \( \times \) and \( \oplus \) acts like \( + \) in the sense that, for example \( L \otimes (M \oplus N) \cong (L \otimes M) \oplus (L \otimes N) \). It is also true that \( L \otimes (M \otimes N) \cong (L \otimes M) \otimes N \).
The tensor product can be used for "extension of scalars": we have a module, say \( \mathbb{Z}^n \) over \( \mathbb{Z} \) and we would like to extend scalars from \( \mathbb{Z} \) to \( \mathbb{Q} \). We can do this by forming the tensor product \( \mathbb{Q} \otimes \mathbb{Z}^n \cong \mathbb{Q}^n \). In many cases this works like you would expect: \( R \otimes R^n \cong R^n \). If \( V \cong F^n \) is an \( F \)-vector space and \( K \supset F \) is a field extension, then \( K \otimes V \cong K^n \). But it is not always so obvious: if \( A \) is a finite abelian group then \( \mathbb{Q} \otimes A \) is zero.
There are many more examples of such facts:
An exact sequence of modules \( L \overset{\phi}{\rightarrow} M \overset{\theta}{\rightarrow} N \) is one for which \( \text{Im}(\phi) = \text{Ker}(\theta) \). This little definition simplifies a lot of arguments. Apparently at the time it was invented it was a signficant advance.
One thing that can be proved with this definition is the Five Lemma, which has even been featured in a movie. The most common application of this lemma is to long exact sequences in (co-)homology.
A projective module is one which is "almost" free, in that it can substitute for a free module in many contexts. There are four equivalent definitions of a module \( P \) being projective:
I suggest trying to memorize some examples of modules that are and are not projective:
An injective module is in a categorical sense the opposite of a projective module. Again, there are various versions of the definition: \( Q \) is an injective module if
Again one should memorize some examples of modules that are and are not injective.
The definition of flat is that tensoring preserves exactness. It's good to have in mind a few facts and examples.
This chapter is a rapid course in linear algebra over an arbitrary field, which should mostly be familiar from your undergraduate course in linear algebra.
This section proves that dimension is well-defined and that finite-dimensional vector spaces are classified (up to isomorphism) by their dimension. The key technical result is a "Replacement Theorem". (This is sometimes called the Steinitz Exchange Lemma.) One way to understand this theorem is that it is the best possible version of a certain type of result. To see this, consider the following lemmas, whose proofs are simple.
(Span Preservation Lemma) If \( \{v_1, \ldots, v_n \} \) is a linearly dependent set of vectors, then there is a \( v_i \) which can be removed without affecting the span.
(Independence Extension Lemma) If \(X= \{v_1, \ldots, v_n \} \) is a linearly independent set of vectors and \( u \) is a vector not in the span of \(X\), then \( X \cup \{ u \} \) is linearly independent.
(Span Preservation Lemma 2) If \(X= \{v_1, \ldots, v_n \} \) is a set of vectors and \( u \neq 0 \) is a vector in the span of \(X\), then there is a \( v_i \) such that \( X - \{v_i\} \cup \{ u \} \) has the same span as \(X\).
Improving the last lemma to handle a set of linearly independent vectors \( \{u_1, \ldots, u_n \} \), we would (eventually) figure out the Replacement Theorem. This theorem says that if we have a basis \(B\) for \(V\) and a linearly independent set \( A \subset V \), we can replace the elements of \( B \) with elements of \( A \) while retaining the basis property.
Once one has proved the replacement result, the rest of the section follows in a formal and trouble-free sort of way. In particular one gets:
Of course linear algebra has many applications, but this section gives one that's important for group theory: the order of the general linear group over a finite field is computed by noting that an invertible \( n \times n \) matrix over \( \mathbb{F}_q \) is (via column vectors) a list of \( n \) linear independent vectors in \( \mathbb{F}_q^n \). The order of the group is therefore \( (q^n - 1)(q^n-q)(q^n-q^2) \cdots (q^n-q^{n-1}) \).
I have been told, and even am willing to believe, that the Steinitz Exchange Lemma is really the same as row reduction. However, I have not written down a correspondence between the two.
The matrix of a linear transformation relative to bases is defined and its properties studied. The fact that matrix multiplication corresponds to composition of linear transformations comes out of this.
The exercises of this section contain a proof of the familiar but not obvious fact that the row rank is the same as the column rank. There is another proof in the next section.
The proof here in outline says: row operations don't change the row rank, because we check it for the 3 elementary row operations. Row operations don't change the set of solutions to linear equations, which are linear dependence relations amongst the columns of the matrix. Therefore row operations don't change the column rank. Thus, we only have to check the equality of row and column rank for the reduced row echelon form. In that form, the nonzero rows are linearly independent, and the pivot columns are linearly independent and form a basis for all the columns. But the number of pivot columns and the number of nonzero rows are equal, by definition of reduced row echelon form.
This section contains another proof that the row and column ranks of a matrix are equal. This is based on several important facts about a finite dimensional vector space \(V \) and its dual \( V^\star = \text{Hom}(V,F) \):
To prove that the row and column ranks are the same it is enough to prove that \( \text{rank}(M(\phi)) = \text{rank}(M(\phi^\star)) \), or equivalently that \( \dim \text{Im}(\phi) = \dim \text{Im}(\phi^\star) \). Apply the rank nullity theorem to \( \phi^\star \) and get \( \dim \text{Im}(\phi^\star) + \dim \text{Ker}(\phi^\star) = \dim W^\star \). Now apply fact 4 to get \( \dim \text{Im}(\phi^\star) + \dim \text{Ann} ( \text{Im} (\phi) ) = \dim W^\star \). Then use fact 3 to get \(\dim \text{Im}(\phi^\star) + \dim W - \dim \text{Im}( \phi) = \dim W^\star \). Then apply fact 1 to cancel \( \dim W \) and \( \dim W^\star \), obtaining the result.
Determinants are familiar from undergraduate linear algebra, but the existence and uniqueness of such an object are not obvious. The definition of a determinant is that it is an alternating multilinear \(n\)-form on \( R^n \), normalized so that the determinant of the identity matrix is 1. Here \( R \) is a ring (taken to be \( \mathbb{R} \) in the undergraduate course).
To prove the existence of a determinant, one can write down the degree \( n \) polynomial in \( n^2 \) variables: \[ \det ( a_{i, j} ) = \sum \epsilon (\sigma) a_{\sigma(1), 1} a_{\sigma(2),2} \cdots a_{\sigma(n), n} \] and show that it has the required properties. The uniqueness comes from showing that the multilinearity allows one to compute the determinant of any matrix from the determinant of the identity matrix (Propositions 22 and 23).
Then several important facts about the determinant are recorded:
We said earlier that if \( V \) was a vector space with basis \( \{e_i\} \) then we could think of \( V \otimes V \) as being a vector space with basis \( \{ e_i \otimes e_j \} \). We can define a product \( V^{\otimes 2} \times V \rightarrow V^{\otimes 3} \) by juxtaposing symbols: \( e_i \otimes e_j \cdot e_k = e_i \otimes e_j \otimes e_k \). Now generalize this idea to \( V^{\otimes l} \times V^{\otimes m} \rightarrow V^{\otimes (l+m)} \).
Having created these products we can create a graded ring \(T(M) = R \oplus M \oplus M^{\otimes 2} \oplus M^{\otimes 3} \oplus \cdots \), which is called the tensor algebra of \(M\). This construction works even if \(M\) is not a vector space, but of course the rules of tensor product may make some of the products zero.
We can add rules requiring this multiplication to be commutative. These rules correspond to taking a quotient of the tensor algebra by an ideal generated by relations, which are elements like \( a \otimes b - b \otimes a \). Having done this we get the symmetric algebra of \(M\). In the simple case, where \( M = V\) is a vector space with basis \( \{ e_i \} \), \(S(M)\) is the polynomial algebra on the variables \( e_i \).
We could instead introduce rules requiring an anti-commutativity. These rules correspond to taking the quotient by the ideal generated by all elements of the form \(m \otimes m\). This produces the exterior algebra \( \Lambda(M) \). In the case where \( M = V \) is a vector space of dimension \( n \), the vector space \( \Lambda^k(M) \) has dimension \( \binom{n}{k} \).
In the case where \( k! \) is invertible in \(R\) it is possible to identify the symmetric and exterior tensors \(S^k(M)\) and \( \Lambda^k(M) \) with subspaces of \(T(M)\), by defining symmetrizing and anti-symmetrizing maps \( \text{Sym}^k \colon T^k(M) \rightarrow S^k(M) \) and \( \text{Alt}^k \colon T^k(M) \rightarrow \Lambda^k(M) \): \[ \text{Sym}^k(z) = \frac{1}{k!} \sum_{\sigma \in S_k} \sigma (z) \qquad \text{Alt}^k(z) = \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma (z) \] Here \( \sigma \) rearranges the elements of simple tensors: \( (12) (a \otimes b) = b \otimes a \).
This chapter continues the course in linear algebra.
The Fundamental Theorem of Finite Abelian Groups is really the Fundamental Theorem of Modules over a PID. Since a vector space with a linear operator \( T \colon V \rightarrow V \) is a module over \( F[x] \), we can apply this theory to vector spaces and obtain the Jordan Canonical Form for matrices. We also obtain the Rational Canonical Form and the Smith Normal Form.
If the field is algebraically closed,