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Download from scratch series........... Maximal Ideal Theorem The quotient of a
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from scratch series........... Maximal Ideal Theorem The quotient of a commutative unital ring by an ideal is a field if and only if the ideal is maximal. BACKGROUND: This is a key theorem that allows the construction of fields and ultimately permits a characterization of all fields by possible orders. KEY DEFINITIONS: 1) A commutative unital ring is a ring with a multiplicative identity and commutative multiplication. 2) A field is an integral domain in which the nonzero elements form a commutative group under multiplication. 3) An ideal is a subring that absorbs all products whenever one factor is in the ideal. In a commutative ring, chirality (handedness) is a moot point, but in noncommutative rings, ideals can be either left or right absorbing (the nonideal factor goes on the left or right, respectively) and then they are called left or right ideals. Note that Rotman’s terminology is complicated by his decreeing that all rings in Chapter 3 shall be commutative and have an identity. Then ideals cannot be subrings, by Rotman’s provisional definition, since once a unit gets into an ideal, it takes over the whole ring. I like the basic definitions that keep the various properties separate and allow us to mix and match to get what we want. Our ideals can be subrings. 4) A quotient ring or factor ring is analogous to a factor group. In fact, as an additive group, a factor ring is exactly a factor group. The additional operation of coset multiplication is defined in the expected way, and coset multiplication distributes over coset addition by virtue of the original ring distributivity. Unlike the situation with factor groups, where normality has to be checked to ensure validity of the coset operation, any additive subgroup in a ring is normal because ring addition is always commutative.Whether a factor ring can be built depends on showing that the additive subgroup is multiplicatively absorbing (a fortiori multiplicatively closed). Thus the subring test is only: for all a, b I R if a b I and ab I, then I is a subring of R. But the ideal test (assuming a commutative ring R) is: for all a, b I R if a b I and ar I, for all r R, then I is an ideal of R. 5) A maximal ideal is one that allows no other proper ideal between it and the full ring. Symbolically, if M R is maximal in the ring R, then if X is another ideal such that M I R, then either M I or I R. 6) The nullring is the ring consisting of only the element 0. Integral domains and fields both have at least two elements, since it is required that 0 1. KEY FACTS: 1) Fields have only the nullring and the entire field as ideals. PROOF STRATEGY: We will use the maximality of the ideal M to explicitly write the multiplicative inverse of an arbitrary nonzero factor ring element. Remember that the factor ring 0 is really M, and the multiplicative identity is the particular coset 1 M. PROOF: (Necessity) Suppose R/M is a field. Fields, by definition, have at least two elements, so M R. If M were not maximal in R, there would be an ideal I strictly between M and R. But then the subfield I/M would be an ideal of R/M. This would force I/M 0 or R/M, due to the triviality of ideals within fields. The first case is impossible because M I., and the second case implies I M, which shows that M was maximal. (Sufficiency) Let M be maximal in R. R/M inherits the operational properties needed for a field from the commutative unital ring R. The only loose end is to show the existence of multiplicative inverses for all nonzero elements. The nonzero elements of R/M are the cosets a M, where a M, and the multiplicative identity is 1 M, so we are done if we can find an element b M such that a Mb M 1 M. Consider the set Ra M. I claim this is an ideal in R. By the ideal test, clearly x, y Ra M forces x y Ra M. Then since mutliplying by an arbitrary x R returns xRa xM Ra M, we see it is an ideal. Now Ra M ra m : r R, m M, so if r 0 we have immediately M Ra M. We conclude that M Ra M R. By the maximality of M, either Ra M M or Ra M R. The former is impossible, since a M, but if we choose r 1 and m 0, a Ra m. Therefore Ra M R, and it must be true that for some b and m, ba m 1. Now a Mb M ab M, and since ba 1 m, a Mb M 1 m M 1 M. We have found a multiplicative inverse for an arbitrary nonzero coset in R/M, and this is enough to conclude that R/M is a field.. APPLICATION: One traditional source of commutative unital rings involves polynomials in (usually) one indeterminate over a field of coefficients. Let us find some maximal ideals to form factor rings with and explore the consequences. But first...here is a lemma that will spit out lots of maximal ideals. An irreducible element in a ring is a nonzero nonunit that forces one of the two factors in a factorization of it to be a unit. Essentially, an irreducible element only admits a trivial factorization. The set of all multiples of px by polynomials in x is clearly an ideal...once a multiple, always a multiple! We denote this by px. A common symbol used to indicate the degree of a polynomial is . LEMMA: Let be a field and px x. Then px is maximal in x if and only if px is irreducible over . PROOF: Suppose that px is maximal in x. I claim px is not a unit. If it were, by the standard argument, px would expand to all of x, and no longer be maximal. Neither is px 0, otherwise px would be the zero ideal in x, and there clearly are ideals strictly between 0 and x, say x or 2, for example. Now suppose px factors as gxhx. Then forming the ideal generated by gx, we would argue that every multiple of px can be achieved by using a suitable multiple of gxhx, hence gx alone, so px gx x. By the assumed maximality of px, we would have either px gx or gx x. If px gx, we can infer that p g and also g p, so g p. This leaves no room for the degree of the companion factor hx to be anything but 0. So hx is a constant, and the nonzero constants in the polynomial ring are precisely the nonzero field elements, hence units, as they are all invertible. So px gx implies a trivial factorization. On the other hand, if gx x, the degree of gx had better be 0, or there would be no way to generate the constants that are in x. Once again the proposed factorization is determined to be trivial. We have shown that the maximality of px in x implies the irreducibility of px over . Conversely, suppose px is irreducible over . We want to show px maximal in x, so to that end suppose there is an ideal I between px and x. A key fact about polynomial rings over fields is that they are always principal ideal domains...that is they have the property that every ideal is generated by a single polynomial (obviously of lowest degree in the ideal). So we may assume the ideal I is generated by some polynomial gx. Then px gx x. We see that px gx, so write px gxhx for some hx x. Now we have assumed that px is irreducible over , so either g or h is zero. If g 0, then 1 gx and gx x. If h 0, then px gx. In either case, we see that px must be maximal. So the hunt is on for irreducible polynomials. Certainly px x 2 1 is famously irreducible over . Then x/x 2 1 is a field. Typical elements of this field are a bx x 2 1, where a, b . There is no point in writing any higher powers of x, since we are saying that every time we see an x 2 we will replace it with 1 so that x 2 1 acts as a zero. Let us do a calculation in this field. a bx x 2 1a bx x 2 1 a 2 b 2 x 2 x 2 1. But b 2 x 2 b 2 x 2 1 b 2 , so by absorption, a 2 b 2 x 2 x 2 1 a 2 b 2 x 2 1 b 2 x 2 1 a 2 b 2 x 2 1. The ideal x 2 1 tags along in these computations as an explicit zero. Any time a multiple of x 2 1 is obtained, you can stuff in the ideal and it disappears. Do you recognize that we have multiplied a complex number by its conjugate to get the sqaure of its modulus? We have discovered that x/x 2 1 . Instead of , let us use the Galois field GF3. Then x 2 1 is still irreducible. How do you know this quickly? Polynomials of degree two or three must have a zero in the field to be reducible...think about the unavoidability of having a linear factor in any proper factorization. With a small field, just test all the field elements in the polynomial. No zero...it’s irreducible. Then GF3x/x 2 1 is a field with a typical element a bx x 2 1 as before. But there are only 3 choices each for a and b, so this field has 9 elements. How would we manufacture a field with q p n elements, where p is a prime? Based on the last example, an approach would be to use GFn.as the field and find an irreducible polynomial pxover GFn.with degree n. Then GFnx/px would be a field, and a typical element might be a 0 a 1 x a n1 x n1 px, and there would be q p n such elements. For example, if we want a representation of a field with 27 elements, we can use GF3 again, but this time choose an irreducible cubic. I think x 3 x 1 will do, as no field elements are zeroes. So GF3x/x 3 x 1 is a field with 27 elements. © 2010 Thomas Beatty