## Polynomial Rings in Relation to Euclidean Domains, PIDs, and UFDs (Pt. I)

**Point of Post:**

** **In this post we discuss how polynomial rings act when their underlying rings are PIDs, Euclidean domains, and UFDs. The high points being the proof that is a Euclidean domain for fields and is a UFD when is a UFD.

*Motivation*

** We have mentioned several times that polynomial rings occupy a central role in modern mathematics. Consequently, studying them should occupy no small amount of our time. In this post we shall prove some previously made assertions about when polynomial rings are Euclidean domains, PIDs, and UFDs and some of the consequences of thes results.**

*Polynomial Rings Over Fields are Euclidean Domains*

** **

We have mentioned several times now that polynomial rings where is a field are Euclidean domains. We now prove this result:

**Theorem: ***Let be a field, then for every there exists unique such that and or .*

**Proof: **If we may take and if then we may take and . So, from now on we assume that and . We proceed then by induction on . If then and the rest follows since is a field. So, assume the result is true for and let , say . Let us be given with, by assumption, . Consider then the polynomial . We know that is of degree less than and so by the induction hypothesis there exists with with or . Letting then gives us that with or . This completes the induction. To see that this is unique suppose that we see then that and since the degree of the right hand side is strictly less than the degree of we must have that and so from where it easily follows that .

With this we can now state:

**Theorem: ***Let be a field, then is a Euclidean domain and satisfies the -inequality.*

This gives us the unproven part of:

**Theorem: ***Let be a ring. Then, is a PID or a Euclidean domain if and only if is a field.*

Now, the mere existence of this division algorithm allows us to conclude some important things about polynomials over a field:

**Theorem: ***Let be a field and . Then, if and only if .*

**Proof: **Evidently if then . Conversely, suppose . We know that where is a constant. Since we may conclude that and so .

From this and the fact that the degree function is additive allows one to conclude the following:

**Corollary: ***Let be a field and be non-zero. Then the number of distinct roots in of does not exceed .*

We are now in position to prove a long-discussed theorem. Namely:

**Theorem: ***Let be an infinite integral domain and the ring of all functions with the usual operations. Then the obvious map is an embedding.*

**Proof: **We have already remarked that this map is a unital homomorphism of rings, and so it suffices to prove that it’s injective. This is equivalent though to showing that if satisfies for all then . To see this we proceed in two steps. First, we prove this result for the field of fractions of . But, this is obvious, if is identically zero as a function on then has infinitely many roots in and so by the previous corollary must be zero.

Suppose now that is identically zero as a function on we claim that it is identically zero as a function on . But, this is obvious since then would have infinitely many roots in . Thus, is the zero polynomial.

So, in an integral domain viewing a polynomial as a set of coefficients and as a function are the same thing.

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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