Introduction to Algebra
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1 MTH4104 Introduction to Algebra Solutions 8 Semester B, 2019 * Question 1 Give an example of a ring that is neither commutative nor a ring with identity Justify your answer You need not give a complete proof, but you should give a a counterexample to the commutative law for multiplication; b an proof that your ring contains no multiplicative identity element; c a general reason for why the ring axioms are true This can be short but, as always, should be in complete sentences Solution There are lots of examples; here s one In lecture I mentioned that an example of a ring without identity is the set of even integers 2Z {2k : k Z} Indeed you could use multiples of any integer greater than 1, not just 2 This doesn t solve the problem by itself, as 2Z is commutative But perhaps we can build a noncommutative ring from it? I also mentioned that matrix rings are usually not commutative, so perhaps a matrix ring over 2Z will give us our example This indeed works To be concrete, let s show that M 2 2Z is a noncommutative ring without identity a A counterexample to the commutative law for multiplication in M 2 2Z is given by 2 0 which is not equal to b Nor does M 2 2Z satisfy the multiplicative identity law Indeed, in any product of two matrices in this ring, say 2a 2b 2e 2 f 4ae + bg 4a f + bh 2c 2d 2g 2h 4ce + dg 4c f + dh 1
2 for integers a,b,c,d,e, f,g,h, every entry of the product is a multiple of 4 Therefore, 2 2 since 2 is not a multiple of 4, there is no matrix in M 2 2Z whose product with will be again 2 2 c M 2 2Z is a ring by facts from lectures, since we argued that 2Z is a ring last week, and M n R is a ring for any ring R and integer n 1 Question 2 a ring This is this week s continue the proofs from lecture question Let R be a Prove that the equation AB+C AB+AC is true for all A,B,C M 2 R This is half of the distributive law It is sometimes called the left distributive law, because the factor being distributed, A, is on the left b Prove the left distributive law for R[x] a11 a Solution a Write A 12 b11 b, B 12 c11 c and C 12 where a 21 a 22 b 21 b 22 c 21 c 22 the coefficients a 11,a 12,a 21,a 22,b 11,b 12,b 21,b 22,c 11,c 12,c 21,c 22 are elements of the ring R b11 + c We have B +C 11 b 12 + c 12 Hence b 21 + c 21 b 22 + c 22 a11 a AB +C 12 b11 + c 11 b 12 + c 12 a 21 a 22 b 21 + c 21 b 22 + c 22 a11 b 11 + c 11 + a 12 b 21 + c 21 a 11 b 12 + c 12 + a 12 b 22 + c 22 a 21 b 11 + c 11 + a 22 b 21 + c 21 a 21 b 12 + c 12 + a 22 b 22 + c 22 The distributive law is valid in the ring R Applying it to each entry in this matrix twice ie expanding all the brackets, whilst being careful not to write say b 21 a 12 instead of a 12 b 21, we get a11 b AB +C 11 + a 11 c 11 + a 12 b 21 + a 12 c 21 a 11 b 12 + a 11 c 12 + a 12 b 22 + a 12 c 22 a 21 b 11 + a 21 c 11 + a 22 b 21 + a 22 c 21 a 21 b 12 + a 21 c 12 + a 22 b 22 + a 22 c 22 On the other hand, we have a11 a AB + AC 12 b11 b 12 a11 a + 12 c11 c 12 a 21 a 22 b 21 b 22 a 21 a 22 c 21 c 22 a11 b 11 + a 12 b 21 a 11 b 12 + a 12 b 22 a11 c a 12 c 21 a 11 c 12 + a 12 c 22 a 21 b 11 + a 22 b 21 a 21 b 12 + a 22 b 22 a 21 c 11 + a 22 c 12 a 21 c 12 + a 22 c 22 a11 b 11 + a 12 b 21 + a 11 c 11 + a 12 c 21 a 11 b 12 + a 12 b 22 + a 11 c 12 + a 12 c 22 a 21 b 11 + a 22 b 21 + a 21 c 11 + a 22 c 21 a 21 b 12 + a 22 b 22 + a 21 c 12 + a 22 c 22 This is equal to AB +C by the commutativity of addition in R 2
3 b Let A, B, and C be three elements of R[x] We may write them out, using sigma notation to be concise, as A m i0 a ix i, B n i0 b ix i, and C n i0 c ix i, where the a i, b i, and c i are elements of R Note that I have taken the same upper bound, n, in the sums for B and C I can do this: if their degrees are actually different, I can append terms with zero coefficients to the front of whichever one has lesser degree I ve done so only for convenience when I write out the sum We must evaluate AB + C and AB + AC, and compare them Now B + C n i0 b i + c i x i, whence m+n AB +C a j b k + c k x i i0 j+ki On the other hand the sum of AB m+n a j b k x i i0 j+ki and is AB + AC AC m+n a j c k x i i0 j+ki m+n a j b k + i0 j+ki a j c k x i j+ki So what is left to do is show that corresponding coefficients of these two polynomials are equal, namely, that a j b k + c k a j b k + a j c k j+ki j+ki j+ki This is true by the ring axioms in R First, using distributivity on each summand of the sum on the left hand side shows it equal to a j b k + a j c k j+ki Now, a succession of uses of associativity to manipulate parentheses suppressed in the notation and commutativity lets us separate the a j b k terms from the a j c k terms, showing this is equal to the right hand side Question 3 Find a multiplicative inverse for the matrix [7] [6] [10] [4] within M 2 Z 3
4 Solution Since Z is a field, we may use the familiar procedure for inversion of 2 2 matrice The rule is that the matrix a b has as its inverse the matrix 1 ad bc c d d b c a supposing that this latter matrix is defined, that is as long as ad bc 0 But it is important that Z be a field for this approach to be correct! Proving it uses not only the multiplicative inverse law, to divide by ad bc, but also less obviously the commutative law In the present instance, we compute that [7] [4] [6] [10] 1 [ 32] 1 [7] 1 [2] so the inverse matrix is [7] [4] [6] [10] 1 [4] [6] [10] [7] [4] [7] [2] [3] [7] [8] [1] [6] [1] The field Z is small enough that brute force is probably the quickest way to observe that [7] 1 [2], but you could also have used the Euclidean algorithm If you were unsure about the applicability of the above procedure, the results could of course be checked, by working out the products [7] [6] [8] [1] [ ] [ ] [1] [0] [10] [4] [6] [1] [ ] [ ] [0] [1] and [8] [1] [7] [6] [ ] [ ] [1] [0] [6] [1] [10] [4] [ ] [ ] [0] [1] What would you do if you didn t know, or sensibly didn t trust, the determinant rule for computing the inverse? In this case, the task would be to solve for elements w,x,y,z Z that bring about [7] [6] w x [1] [0] [10] [4] y z [0] [1] and w x [7] [6] [1] [0] y z [10] [4] [0] [1] Performing either one of these multiplications and equating entries gives you two lots of two linear equations in two unknowns: if we use the first, we get [7] w + [6] y [7] x + [6] z [1] [0] [10] w + [4] y [10] x + [4] z [0] [1] 4
5 which implies [7] w + [6] y [1] [10] w + [4] y [0] ; [7] x + [6] z [0] [10] x + [4] z [1] These can be solved by the usual isolate-and-substitute procedure, and produce the same answer as above Question 4 Let K be a skewfield, and let f,g K[x] be two nonzero polynomials Prove that f g 0 Solution Suppose that f has degree d and g has degree e Since neither f nor g are zero, both of these degrees are well-defined It follows that f a d x d + + a 1 x + a 0 g b e x e + + b 1 x + b 0 where a d and b e are nonzero elements of R Now, the product of f and g is the sum of a collection of terms like a i b j x i+ j The only way the exponent i + j of x in this term can be as large as d +e is if i d and j e, which are the largest possible values of i and j Therefore f g a d b e x d+e + terms with smaller powers of x; in other words, if we write f g c n x n + + c 1 x + c 0 in standard for, then n d + e and c n a d b e It is true in any skewfield that if a d 0 and b e 0 then a d b e 0 This needs proof! A quick way to show it is that a d b e has a multiplicative inverse, namely b 1 e a 1 d, while 0 cannot possibly have a multiplicative inverse Having established this, it follows that the leading coefficient of f g is nonzero, and therefore f g is not the zero polynomial, as desired Question 5 Suppose R is a nontrivial ring with identity a Specify a non-zero 2 2 matrix N with coefficients in R such that N 2 0 b Using part a or otherwise, prove that M 2 R does not satisfy the multiplicative inverse law 5
6 Solution a Clearly if we take the zero matrix O O 2 OO, then O But we need a non-zero matrix So let s try changing just one of the zeros in O to a non-zero entry, let s say the multiplicative identity 1 R, which we know is not zero because R satisfies the nontriviality law 1 0 The matrix doesn t work since it is its own square check this for yourself! For similar reasons, doesn t work either However 0 1 does work, because N N : O b We know that M 2 R is a ring from the lectures Hence we may apply our general argument about nonzero elements x N, y N in M 2 R that satisfy xy 0 inside M 2 R, to deduce that M 2 R is not a field To wit, if x had a multiplicative inverse x 1, then y 1 y x 1 xy x which is false Question 6 Let R be a ring, and n a natural number Describe the rings M n R[x] and M n R[x] Explain how these two rings relate to each other Solution M n R[x] is the ring of n-by-n matrices whose entries are polynomials in the variable x with coefficients in R, whereas M n R[x] is the ring of polynomials in the variable x whose coefficients are n-by-n matrices with entries in R For example, when n 2 and R Z, an element of each ring is given by 1 x 1 M x 1 2 Z[x], x + M Z[x] The relationship between these rings is one that should be becoming a familiar theme: they are two different presentations of the same ring You can go back and forth between them Given a polynomial of matrices, you can formally multiply the powers of x through inside the matrices, sum the results, and end up with a matrix of polynomials To go the other direction, given a matrix of polynomials over R, you can 6
7 split it as a sum of matrices one of which contains only elements of R, one only elements of R times x, one only elements of R times x 2, etcetera, and then move the power-of-x factors outside The two examples above are related by this procedure But this is not enough to make the rings M n R[x] and M n R[x] the same, since a ring is not just its set of elements; it also has rules for addition and multiplication So the key facts are that, if you are adding or multiplying two matrices of polynomials, you d get the same results if you instead translated to polynomials with matrix coefficients and worked the sum or product out in that language instead The proofs of these facts are, like many others in the past week, not difficult but tedious, so I ve omitted them Formally, this is yet another isomorphism, like those we saw on the previous coursework Question 7 Let R be a ring Prove that the set { } a b : a,b,c R 0 c of 2 2 matrices over R whose lower left entry equals zero is a ring, with the usual addition and multiplication of matrices [You may assume that M 2 R is a ring] Solution axioms Let S be the set in the question We will show S is a ring by proving the ring The key feature of our situation is that S is defined as a subset of the ring M 2 R As emphasised in lecture, this is a situation where the closure laws are particularly important On the other hand, a number of the ring axioms for S will automatically be true by virtue of the fact that elements of S are elements of M 2 R, and the operations work the same way For brevity, I will just say automatic below for these laws Closure for + We must show that the sum a b d e + 0 c 0 f of two elements of S is also in S This sum is a + d b + e, 0 c + f which has the requisite zero in the lower-left corner, and is therefore in S Closure for Now we must compute the product a b d e 0 c 0 f and show that it is in S The product is a d + b 0 a e + b f 0 d + c e + c f ad ae + b f, 0 c f which again is in S by virtue of the lower-left entry being zero 7
8 Associativity for + Automatic Associativity for Automatic Identity for + We note that the additive identity matrix of M 2 R is in the set S Then the fact that it still behaves as an identity for the addition in S is automatic Inverses for + Given a matrix a b 0 c in S, the matrix a b, 0 c which is its additive inverse within M 2 R, lies in S Then the fact that it s still the inverse in S is automatic Commutativity for + Automatic Distributivity Automatic 8
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