Download from scratch series........... Maximal Ideal Theorem The quotient of a

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  Mb  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  Mb  M  ab  M,
and since ba  1  m, a  Mb  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 px by polynomials in x is clearly an ideal...once a multiple, always a multiple! We
denote this by px. A common symbol used to indicate the degree of a polynomial is .
LEMMA:
Let  be a field and px  x. Then px is maximal in x if and only if px is irreducible
over .
PROOF:
Suppose that px is maximal in x. I claim px is not a unit. If it were, by the standard
argument, px would expand to all of x, and no longer be maximal. Neither is px  0,
otherwise px 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 px factors as gxhx. Then forming the ideal
generated by gx, we would argue that every multiple of px can be achieved by using a suitable
multiple of gxhx, hence gx alone, so px  gx  x. By the assumed maximality of
px, we would have either px  gx or gx  x. If px  gx, we can infer that
p  g and also g  p, so g  p. This leaves no room for the degree of the companion factor
hx to be anything but 0. So hx 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 px  gx
implies a trivial factorization. On the other hand, if gx  x, the degree of gx 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 px in x implies
the irreducibility of px over .
Conversely, suppose px is irreducible over . We want to show px maximal in x, so to
that end suppose there is an ideal I between px 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 gx. Then px  gx  x. We see that
px  gx, so write px  gxhx for some hx  x. Now we have assumed that px is
irreducible over , so either g or h is zero. If g  0, then 1  gx and gx  x. If h  0,
then px  gx. In either case, we see that px must be maximal. 
So the hunt is on for irreducible polynomials. Certainly px  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  1a  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 GF3. 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 GF3x/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 GFn.as the field and find an irreducible polynomial
pxover GFn.with degree n. Then GFnx/px would be a field, and a typical element might
be a 0  a 1 x    a n1 x n1  px, 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 GF3 again, but this time choose an
irreducible cubic. I think x 3  x  1 will do, as no field elements are zeroes. So GF3x/x 3  x  1
is a field with 27 elements.
© 2010 Thomas Beatty