Suppose is an integral domain. There is always a ring homomorphism from that sends where is the identity element in .
By the isomorphism theorem, this gives an injective map where is the kernel of .
Since the image of this map is an integral domain, is a prime ideal in . Therefore either or for some prime . In the first case we say that has characteristic zero and in the second we say that has characteristic .
In the first case, contains a copy of ; in the second, a copy of .
In an integral domain of characteristic , for any . In characteristic zero, if for and , then .
The rings , , , , , all have characteristic zero.
The rings , , have characteristic .
Lemma: In a ring of characteristic , .
Proof: The binomial coefficients are divisible by whenever .
Fraction fields
Suppose that is an integral domain. We can construct a field containing considering and imposing the usual “fraction rules”:
if .
More formally we can consider ordered pairs with and define an equivalence relation saying for all in ; then defining the operations as above on equivalence classes, checking everything is well defined, and so on. See DF Section 7.5, especially Theorem 15 of that chapter, for all the details.
Proposition: Suppose that is an integral domain and is a field. If is an injective map sending to , then extends to a map from into . Expressed differently, if is a field that contains a subring isomorphic to , then the smallest subfield of containing is isomorphic to . Informally, is the smallest field containing .
Examples:
.
= (this is notation) – the field of rational functions over .
= – the field of rational functions over .
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Polynomial Rings
Let be a commutative ring with unity.
An element is monic if its highest degree coefficient is .
The units in are the units in .
If is an integral domain, so is (look at highest degree terms of the polynomials)
If is an ideal of , then is isomorphic to .
If is a prime ideal in , then is a prime ideal in .
If is a monic polynomial in and is any polynomial, then there is a division algorithm yielding with the degree of less than the degree of .
If is a field, any polynomial can be made monic multiplying by the inverse of its highest degree coefficient.
The ring is the ring of polynomials in variables with coefficients in . The terms of such a polynomial are monomials The total degree of such a monomial is the sum of its degrees, and the total degree of a polynomial is the highest total degree of its monomials.
A polynomial in may also be viewed as a polynomial in whose coefficients are polynomials in . (In other words, ).x_n$ with nonzero coefficient.
A polynomial in variables is homogeneous if all monomials have the same total degree. Any polynomial in variables can be written as a sum of homogeneous polynomials.
Proposition: is a PID if and only if is a field.
Proof: If is a field we know this by Euclid’s algorithm and polynomial division. If is not a field, it has a proper maximal ideal . Then is prime since is an integral domain. But is not a field – it contains the proper ideal generated by . Thus contains non-maximal prime ideals, so it can’t be a PID.
Polynomial rings over UFD’s are UFD’s.
If is a UFD, a polynomial in is called primitive if the greatest common divisor of its coefficients is ; or, put another way, if the ideal generated by the coefficients in is .
Theorem: (DF, Theorem 7, page 304) is a UFD if and only if is a UFD.
Since is a subring of and a factorization of an element of in involves only elements of , if is a UFD, so is .
Going the other way is the interesting part.
The basic strategy is:
start with a polynomial .
factor out the gcd of the coefficients in so that and is primitive.
The term has a unique factorization, since it’s in , so we have to worry about the primitive polynomial .
View as a polynomial in , which is a polynomial ring over a field and therefore a PID/UFD.
Since the leading coefficient of is a unit in , we can factor it out and write where is a monic polynomial in .
Thus where the are irreducible monic polynomials with coefficients in .
Therefore factors as .
Problem 1: At this point, the have coefficients in – they have “denominators”. We need to clear out those denominators. We have the leading coefficient . Can we somehow split up into pieces so that belongs to ? If so, then we’ve factored where each factor is in and is irreducible in .
Suppose we can do this!
We’re happy because we’ve factored as a product of polynomials in , each of which is irreducible in . And they must also be irreducible in , because .
Problem 2: We need to establish uniqueness. The idea is that if we had two factorizations of in , we’d have two factorizations in . Since is a PID, we can show that these two factorizations are “the same” in – they have the same number of terms, and the terms can be matched up as associates in . But two irreducibles can be associates in and maybe not in – this is another “denominator problem.” So we need to sort that out as well. This reduces to the question: suppose and are irreducibles in that are associates. This means that we can “clear denominators” in and so that both are in and primitive and for some in . Since for elements and in , this amounts to the equation . But since and are primitive, the gcd of the coefficients of is , and of is , so and are equal up to a unit in and and are associates in .