MHB Polynomial Rings - Gauss's Lemma

[Math Amateur]

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) so by Proposition 2 above 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



Note: Proposition 2 referred to above states the following:

Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (the set of polynomials with co-efficients in I). Then

$$ R[x]/(I) \cong (R/I)[x]$$

In particular, if I is a prime ideal of R then (I) is a prime ideal of R[x]

 

[Math Amateur]

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) so by Proposition 2 above 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



Note: Proposition 2 referred to above states the following:

Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (the set of polynomials with co-efficients in I). Then

$$ R[x]/(I) \cong (R/I)[x]$$

In particular, if I is a prime ideal of R then (I) is a prime ideal of R[x]

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

 

[Opalg]

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! :-)

That is correct. In fact, the reason prime ideals were given that name is that they are the ideals generated by primes!

 

[Math Amateur]

That is correct. In fact, the reason prime ideals were given that name is that they are the ideals generated by primes!

Thanks Opalg

Can someone now help with problem (2) above

To repeat this problem is as follows:

(2) Despite reading Proposition 2 in Section 9.1, I cannot see why the ideal $$ p_1 R[x] $$ is prime in R[x] and [FONT=MathJax_Main]$$ (R/p_1 R) [x] $$ 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?

Peter

 

[Opalg]

I cannot see why the ideal $$ p_1 R[x] $$ is prime in R[x] and $$ (R/p_1 R) [x] $$ 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 will write $p$ rather than $p_1$, to avoid tiresome subscripts.

$pR[x]$ is the set of all polynomials whose coefficients are all multiples of $p$. It should be pretty obvious that this is an ideal in $R[x]$. In fact, if $pf(x)\in pR[x]$ and $g(x)\in R[x]$, then $(pf(x))g(x) = p(f(x)g(x))\in pR[x]$.

To see that $pR[x]$ is a prime ideal, suppose that $f(x),\ g(x) \in R[x]$ and $f(x)g(x)\in pR[x]$. We want to show that either $f(x)$ or $g(x)$ is in $pR[x]$. Suppose not: then both $f(x)$ and $g(x)$ have at least one coefficient that is not a multiple of $p$. Writing $f(x) = a_0+a_1x+a_2x^2+\ldots$ and $g(x) = b_0+b_1x+b_2x^2+\ldots$, let $a_m$ be the first coefficient in the expression for $f(x)$ that is​ not a multiple of $p$; and similarly let $b_n$ be the first coefficient in the expression for $g(x)$ that is​ not a multiple of $p$. Then you can check that the coefficient of $x^{m+n}$ in $f(x)g(x)$ is not a multiple of $p$ and hence $f(x)g(x)\notin pR[x]$ – contradiction.

To see that $(R/pR)[x]$ is an integral domain, we must show that it has no zero-divisors. So suppose that $f(x),\ g(x)\in (R/pR)[x]$ with $f(x)g(x)=0$. Because the coefficient ring $R/pR$ has no zero-divisors, it follows (by considering the term of highest degree in the product) that the degree of $f(x)g(x)$ is $\deg(f) + \deg(g)$, which must therefore be 0. But then $f(x)$ and $g(x)$ are both constants, and (again because $R/pR$ has no zero-divisors) one of them must be $0$.