# Thread: Polynomial Rings - Gauss's Lemma

1. ## Polynomial Rings - Gauss's Lemma

I am trying to understand the proof of Gauss's Lemma as given in Dummit and Foote Section 9.3 pages 303-304 (see attached)

On page 304, part way through the proof, D&F write:

"Assume d is not a unit (in R) and write d as a product of irreducibles in R, say $d = p_1p_2 ... p_n$ . Since $p_1$ is irreducible in R, the ideal $(p_1)$ is prime (cf Proposition 12, Section 8.3 - see attached) so by Proposition 2 above (see attached) the ideal $p_1R[x]$ is prime in R[x] and $(R/p_1R)[x]$ is an integral domain. ..."

My problems with the D&F statement above are as follows:

(1) I cannot see why the ideal $(p_1)$ is a prime ideal. Certainly Proposition 12 states that "In a UFD a non-zero element is prime if and only if it is irreducible" so this means $p_1$ is prime since we were given that it was irreducible. But does that make the principal ideal $(p_1)$ a prime ideal? I am not sure! Can anyone show rigorously that $(p_1)$ a prime ideal?

(2) Despite reading Proposition 12 in Section 8.3 I cannot see why the ideal $p_1R[x]$ is prime in R[x] and $(R/p_1R)[x]$ is an integral domain. ...". (Indeed, I am unsure that $p_1R[x]$ is an ideal!) Can anyone show explicitly and rigorously why this is true?

I would really appreciate clarification of the above matters.

Peter

2. ## Re: Polynomial Rings - Gauss's Lemma

In trying to answer my problem (1) above - I cannot see why the ideal $(p_1)$ is a prime ideal - I was looking at definitions of prime ideals and trying to reason from there.

I just looked up the definition of a prime element in D&F to find the following on page 284:

The non-zero element $p \in R$ is called prime if the ideal (p) generated by p is a prime ideal!

So the answer to my question seems obvious:

$p_1 irreducible \Longrightarrow p_1$ prime $\Longrightarrow (p_1)$ prime ideal

Although this now seems obvious, I would like someone to confirm my reasoning (which as I said now seems blindingly obvious! :-)

Peter

3. ## Re: Polynomial Rings - Gauss's Lemma

If we have $f \in K[x]$ for some polynomial ring $K[x]$ then the following are equivalent
1) f is irreducible
2) (f) is prime
3) (f) is maximal

$\\ Proof: (3) \Rightarrow (2) \Rightarrow (1) \Rightarrow (3) \\ \\ (3) \Rightarrow (2) is obvious by definition of maximal and prime ideals. \\ \\ (2) \Rightarrow (1) \\ Suppose f is not irreducible. \\ Then f=g \cdotp h with deg(g),deg(h) < deg(f) . \\ Now we get g,h are not in (f) but g \cdotp h is in (f). \\ This is a contradiction to (f) being a prime ideal. \\ \\ (1) \Rightarrow (3) \\ We do this by showing if J \supset (f) then J contains a unit. \\ Let J be generated by a single element, say g, so J=(g) . \\ If f \in (g) then f = q \cdotp g for some poly q. \\ f is irreducible so either q or g is a unit. \\ Suppose q is a unit. \\ Then g = q^{-1} \cdotp f \Rightarrow g \in (f) \Rightarrow (g)=(f) . \\ This contradicts J \supset (f) . \\ Thus g is a unit.$

4. ## Re: Polynomial Rings - Gauss's Lemma

You can't say a polynomial itself is prime (at least to my knowledge).
The idea of a prime ideal as far as I know comes from the ideals of the integers generated by a prime number.
But a prime number and an irreducible polynomial are somewhat related, neither can be factored in their given ring/field.

5. ## Re: Polynomial Rings - Gauss's Lemma

Sorry about the piecewise answers but I am doing one part at a time haha.

$p_{1}R[x] is just the ideal (p_{1}) as for an ideal if r \in R and i \in I then r \cdotp i \in I$

6. ## Re: Polynomial Rings - Gauss's Lemma

$\\ R/I is an integral domain \Leftrightarrow I is prime. \\ \\ Proof \\ \\ \Rightarrow \\ Given a \cdotp b \in I we need to show a \in I or b \in I. \\ \bar{a} \cdotp \bar{b} = 0 \ \Rightarrow \ \bar{a} = 0 or \bar{b} = 0 \ \Rightarrow a \in I or b \in I. \\ \\ \Leftarrow \\ Given a \cdotp b \in I we must have \bar{a} \cdotp \bar{b} = 0 as I is the kernel of the canonical homomorphism thus \bar{a} = 0 or \bar{b} = 0.$

7. ## Re: Polynomial Rings - Gauss's Lemma

Originally Posted by rushton
The idea of a prime ideal as far as I know comes from the ideals of the integers generated by a prime number.
Not quite. For example, $(2)$ and $(3)$ are not actually prime in the ring of integers $\mathbb{Z}[\sqrt{-5}]$

The idea of a prime ideal comes from attempts to extend the fundamental theorem of arithmetic. In the same way we have unique (up to sign) prime factorisations of integers in rationals, we want to have to have some sort of phenomena in the ring of integers in other fields, particularly imaginary quadratic fields. However, in the ring of integers $\mathbb{Z}[\sqrt{-5}]$ of $\mathbb{Q}[\sqrt{-5}]$, we have $2\cdot 3 =6= (1+\sqrt{-5})(1-\sqrt{-5})$, so factorisation is certainly not unique. However, setting $A=(2,1+\sqrt{-5}), \overline{A}=(2,1-\sqrt{-5}),B=(3,1+\sqrt{-5}),\overline{B}=(3,1-\sqrt{-5})$, we have

$(6)=(2)(3)=(\overline{A}A)(\overline{B}B)=( \overline{A} \overline{B})(AB)=(1-\sqrt{-5})(1+\sqrt{-5})$.

So in this case, $(2)$ and $(3)$ are not actually prime. It can be shown that the prime 'factors' of the ideal generated by 6 are $A,\overline{A},B,\overline{B}$

It turns out that there is unique factorisation of prime ideals in the ring of integers in any imaginary quadratic field.

8. ## Re: Polynomial Rings - Gauss's Lemma

Yeah but technically isn't $\mathbb{Z} [ \sqrt{-5}]$ the ring of polynomials with coefficients in $\mathbb{Z}$ adjoining $\sqrt{-5}$? So it wouldnt actually be the field of integers?

9. ## Re: Polynomial Rings - Gauss's Lemma

Originally Posted by rushton
Yeah but technically isn't $\mathbb{Z} [ \sqrt{-5}]$ the ring of polynomials with coefficients in $\mathbb{Z}$ adjoining $\sqrt{-5}$? So it wouldnt actually be the field of integers?
$\mathbb{Z} [ \sqrt{-5}]$ is the ring of algebraic integers in the field $\mathbb{Q} [ \sqrt{-5}]$.

10. ## Re: Polynomial Rings - Gauss's Lemma

Ah I get what you mean now, yeah your totally right.

field of integers .......lol

11. ## Re: Polynomial Rings - Gauss's Lemma

Thanks Rushton

You write:

"You can't say a polynomial itself is prime (at least to my knowledge).
The idea of a prime ideal as far as I know comes from the ideals of the integers generated by a prime number.
But a prime number and an irreducible polynomial are somewhat related, neither can be factored in their given ring/field. "

Dummit and Foote on page 284 give the following definitions of irreducible and prime for integral domains.

-------------------------------------------------------------------------------------------------------------------------------
"Definition Let R be an integral domain.

(1) Suppose $r \in R$ is non-zero and not a unit. Then r is called irreducible in R if whenever r = ab with $a, b \in R$ at least one of a or b must be a unit in R. Otherwise r is said to be reducible.

(2) The non-zero element $p \in R$ is called prime in R if the ideal (p) generated by p is a prime ideal. In other words, a non-zero element p is a prime if it is not a unit and whenever p|ab for any $a,b \in R$, then either p|a or p|b."

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So where a ring of polynomials is an integral domain we have a definition of prime and irreducible elements (polynomials). Do you agree? What do you think?

Mind you most algebra books I have referenced just talk about irreducible polynomials - so maybe for polynomials (for some reason) irreducible and prime are the same thing? Can someone clarify this point?

Another point is that I am unsure why D&F restrict these definitions to an integral domain thus leaving the terms undefined for general rings that are not integral domains. Can someone clarify?

Yet another problem I have with the above definitions by D&F is the following: D&F write: "In other words, a non-zero element p is a prime if it is not a unit and whenever p|ab for any $a,b \in R$, then either p|a or p|b." - How does this follow from (p) being a prime ideal.

Peter

Note: D&F's definition of prime ideal is on page 255 and is as follows:

Definition: Assume R is commutative. An ideal P is called a prime ideal if $P \ne R$ and whenever the product of two elements $a,b \in R$ is an element of P, then at least on of a and b is an element of P.