Polynomial Rings - Irreducibility - Proof of Eisenstein's Criteria

[Math Amateur]

I am reading Dummit and Foote Section 9.4 Irreducibility Criteria. In particular I am struggling to follow the proof of Eisenstein's Criteria (pages 309-310 - see attached).

Eisenstein's Criterion is stated in Dummit and Foote as follows: (see attachment)

Proposition 13 (Eisenstein's Criterion) Let P be a prime ideal of the integral domain R and let

f(x) = x^n + a_{n-1}x^{n-1} + ... ... + a_1x + a_0

be a polynomial in R[x] (here n \ge 1 )

Suppose a_{n-1}, ... ... a_1, a_0 are all elements of P and suppose a_0 is not an element of P^2.

Then f(x) is irreducible in R[x]The proof begins as follows: (see attachment)

Proof: Suppose f(x) were reducible, say f(x) = a(x)b(x) in R[x] where a(x) and b(x) are nonconstant polynomials.

Reducing the equation modulo P and using the assumptions on the coefficients of f(x) we obtain the equation x^n = \overline{a(x)b(x)} in (R/P)[x] where the bar denotes polynomials with coefficients reduced modulo P. Since P is a prime ideal, R/P is an integral domain, and it follows that \overline{a(x)} and \overline{b(x)} have zero constant term i.e. the constant terms of both a(x) and b(x) are elements of P. But then the constant term a_0 of f(x) as the product of these two would be an element of P^2, a contradiction.Questions/Issues

(1) I cannot see exactly how the following follows:

"Reducing the equation modulo P and using the assumptions on the coefficients of f(x) we obtain the equation x^n = \overline{a(x)b(x)} in (R/P)[x]"

I am struggling to see just exactly how this follows: can someone please clarify this?

(2) I am somewhat unsure of the following:

"t follows that \overline{a(x)} and \overline{b(x)} have zero constant term"

This probably is a consequence of the mechanics of the reduction modulo P process but can someone please clarify exactly how it follows?

Peter

[NOTE: This has also been posted on MHF]

 

[Opalg]

Let P be a prime ideal of the integral domain R and let

f(x) = x^n + a_{n-1}x^{n-1} + ... ... + a_1x + a_0

be a polynomial in R[x] (here n \ge 1 )

Suppose a_{n-1}, ... ... a_1, a_0 are all elements of P and suppose a_0 is not an element of P^2.

Then f(x) is irreducible in R[x]The proof begins as follows: (see attachment)

Proof: Suppose f(x) were reducible, say f(x) = a(x)b(x) in R[x] where a(x) and b(x) are nonconstant polynomials.

Reducing the equation modulo P and using the assumptions on the coefficients of f(x) we obtain the equation x^n = \overline{a(x)b(x)} in (R/P)[x] where the bar denotes polynomials with coefficients reduced modulo P. Since P is a prime ideal, R/P is an integral domain, and it follows that \overline{a(x)} and \overline{b(x)} have zero constant term i.e. the constant terms of both a(x) and b(x) are elements of P. But then the constant term a_0 of f(x) as the product of these two would be an element of P^2, a contradiction.Questions/Issues

(1) I cannot see exactly how the following follows:

"Reducing the equation modulo P and using the assumptions on the coefficients of f(x) we obtain the equation x^n = \overline{a(x)b(x)} in (R/P)[x]"

I am struggling to see just exactly how this follows: can someone please clarify this?

You are given that $f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$. When you go to the quotient domain $R/P$, this becomes $\overline{f(x)} = x^n + \overline{a_{n-1}}x^{n-1} + \ldots + \overline{a_1}x + \overline{a_0}$. But each of those coefficients $a_j$ is an element of $P$, and hence $\overline{a_j} = 0$. Therefore $\overline{f(x)} = x^n.$

(2) I am somewhat unsure of the following:

"It follows that \overline{a(x)} and \overline{b(x)} have zero constant term"

This probably is a consequence of the mechanics of the reduction modulo P process but can someone please clarify exactly how it follows?

This takes a bit more work. Suppose that $\overline{a(x)} = x^k + p_{k-1}x^{k-1} + \ldots + p_0$ and $ \overline{b(x)} = x^l + q_{l-1}x^{l-1} + \ldots + q_0$ (where all the coefficients are in $R/P$). Then the equation $x^n = \overline{a(x)b(x)}$ becomes $$x^n = \bigl(x^k + p_{k-1}x^{k-1} + \ldots + p_0\bigr) \bigl(x^l + q_{l-1}x^{l-1} + \ldots + q_0\bigr).\qquad(*)$$ Comparing the constant terms on both sides, you see that $0 = p_0q_0.$ But $R/P$ is an integral domain and so has no zero-divisors. Thus at least one of $p_0$ and $q_0$ must be $0$, say $p_0=0$. Maybe some of the other coefficients in $\overline{a(x)}$ are also $0$, so let $p_rx^r$ be the lowest-degree term in $\overline{a(x)}$ where the coefficient is nonzero. Now compare the coefficients of $x^r$ on both sides of $(*)$ to see that $0 = p_rq_0$. But $p_r\ne0$ and hence $q_0=0.$ thus both $p_0$ and $q_0$ are zero, which is what we wanted to prove.

 

[Math Amateur]

You are given that $f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$. When you go to the quotient domain $R/P$, this becomes $\overline{f(x)} = x^n + \overline{a_{n-1}}x^{n-1} + \ldots + \overline{a_1}x + \overline{a_0}$. But each of those coefficients $a_j$ is an element of $P$, and hence $\overline{a_j} = 0$. Therefore $\overline{f(x)} = x^n.$This takes a bit more work. Suppose that $\overline{a(x)} = x^k + p_{k-1}x^{k-1} + \ldots + p_0$ and $ \overline{b(x)} = x^l + q_{l-1}x^{l-1} + \ldots + q_0$ (where all the coefficients are in $R/P$). Then the equation $x^n = \overline{a(x)b(x)}$ becomes $$x^n = \bigl(x^k + p_{k-1}x^{k-1} + \ldots + p_0\bigr) \bigl(x^l + q_{l-1}x^{l-1} + \ldots + q_0\bigr).\qquad(*)$$ Comparing the constant terms on both sides, you see that $0 = p_0q_0.$ But $R/P$ is an integral domain and so has no zero-divisors. Thus at least one of $p_0$ and $q_0$ must be $0$, say $p_0=0$. Maybe some of the other coefficients in $\overline{a(x)}$ are also $0$, so let $p_rx^r$ be the lowest-degree term in $\overline{a(x)}$ where the coefficient is nonzero. Now compare the coefficients of $x^r$ on both sides of $(*)$ to see that $0 = p_rq_0$. But $p_r\ne0$ and hence $q_0=0.$ thus both $p_0$ and $q_0$ are zero, which is what we wanted to prove.

Thanks for that really helpful post Opalg

I have just finished replying to your reply to my post "Specific example of Eisenstein's Theorem using R = Z" and you may have answered my question in that reply.

Basically I am seeking to understand how can we reduce the equation f(x) = a(x)b(x) which is in R[x] by an ideal P which is not even in R[x]?

You mention "When you go to the quotient domain $R/P$, this becomes $\overline{f(x)} = x^n + \overline{a_{n-1}}x^{n-1} + \ldots + \overline{a_1}x + \overline{a_0}$."

But what is involved in "going to the quotient domain" and how/why is the process valid.

Is it indeed using what Dummit and Foote have named "Proposition 2" of Chapter 9 on polynomial rings (given on page 296 - see attached) which states the following:

--------------------------------------------------------------------------------

Proposition 2. Let I be an ideal of the ring R and let (I) = I[x] denote the idea of R[x] generated by I (the set of polynomials with coefficients 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}

----------------------------------------------------------------------------------

But surely if D&F are indeed using this they should talk about reducing the equation f(x) = a(x)b(x) modulo (P) = P[x] not modulo P.

Can someone please clarify the issues involved here?

Peter

 

[Opalg]

I am seeking to understand how can we reduce the equation f(x) = a(x)b(x) which is in R[x] by an ideal P which is not even in R[x]?

You mention "When you go to the quotient domain $R/P$, this becomes $\overline{f(x)} = x^n + \overline{a_{n-1}}x^{n-1} + \ldots + \overline{a_1}x + \overline{a_0}$."

But what is involved in "going to the quotient domain" and how/why is the process valid.

Is it indeed using what Dummit and Foote have named "Proposition 2" of Chapter 9 on polynomial rings (given on page 296 - see attached) which states the following:

--------------------------------------------------------------------------------

Proposition 2. Let I be an ideal of the ring R and let (I) = I[x] denote the idea of R[x] generated by I (the set of polynomials with coefficients 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]

----------------------------------------------------------------------------------

But surely if D&F are indeed using this they should talk about reducing the equation f(x) = a(x)b(x) modulo (P) = P[x] not modulo P.

As far as I can see, you don't actually need D&F's Proposition 2 here. All you need is the fact that the quotient map $a\mapsto \overline{a}:R\to R/P$ induces a natural map from $R[x]$ to $(R/P)[x]$, namely the map that takes $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x+a_0$ to $\overline{a_n}x^n + \overline{a_{n-1}}x^{n-1} + \ldots + \overline{a_1}x+\overline{a_0}$. The fact that $(R/P)[x]$ is isomorphic to $R[x]/(P)$ is essential in proving some results, but I don't think that it is relevant for Eisenstein's Criterion.