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

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

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

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

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.

 

[blue_raver22]

(1) To understand how the equation x^n = \overline{a(x)b(x)} is obtained, we need to first understand what it means to reduce a polynomial modulo a prime ideal P. This means that we replace each coefficient of the polynomial with its residue modulo P. In other words, we take the remainder of each coefficient when divided by P.

Now, since we are assuming that all the coefficients of f(x) are in P, when we reduce them modulo P, we get 0. Therefore, the polynomial \overline{f(x)} becomes x^n, since all the other terms become 0. Similarly, when we reduce the polynomials a(x) and b(x) modulo P, their constant terms become 0, since they are in P. Therefore, we have x^n = \overline{a(x)b(x)} in (R/P)[x], where (R/P)[x] is the ring of polynomials with coefficients in the quotient ring R/P.

(2) The fact that \overline{a(x)} and \overline{b(x)} have zero constant term follows from the fact that the constant terms of a(x) and b(x) are in P. We know that P is a prime ideal, which means that it is closed under multiplication. Therefore, if the constant terms of a(x) and b(x) are in P, then their product, which is a_0, must also be in P. But we are given that a_0 is not in P^2, which means that it cannot be in P. This is a contradiction, and thus it follows that the constant terms of a(x) and b(x) must be 0.