MHB Polynomial Rings - Irreducibility

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently focused on Section 3.7 Irreducibility...

I need help with an aspect of the proof of Theorem 3.97.

Theorem 3.97 and its proof read as follows:
View attachment 4685
Now, the first part of the proof of Theorem 3.97 relies on Theorem 3.3 ... so I am providing the statement of Theorem 3.3 as follows: ... ...https://www.physicsforums.com/attachments/4680
Now the first line of Theorem 3.97 reads as follows:

"By Theorem 3.33, the natural map $$\phi : \ \mathbb{Z} \rightarrow \mathbb{F}_p$$ defines a homomorphism $$\phi^* : \ \mathbb{Z} [x] \rightarrow \mathbb{F}_p [x] $$... ... "

But, in Theorem 3.33 we have that

$$\phi : \ R \rightarrow S $$

and

$$\tilde{ \phi }: R[x_1, x_2, \ ... \ x_n ] \rightarrow S $$

... so both codomains are $$S$$ ...

BUT ... ... This does not match Theorem 3.97 which has the following functions ...

$$\phi : \ \mathbb{Z} \rightarrow \mathbb{F}_p$$

$$ \phi^* : \ \mathbb{Z} [x] \rightarrow \mathbb{F}_p [x] $$

Now the codomains of $$\phi $$ and $$\phi^* $$ should be the same ( $$\equiv S$$ ), but the codomain of $$\phi $$ is $$\mathbb{F}_p$$, and the codomain of $$\phi^* $$ is $$\mathbb{F}_p [x]$$ .

So, then, how exactly is Rotman applying Theorem 3.33 in this context.

Further ... ... it seems the codomain of $$\phi $$ cannot be $$\mathbb{F}_p $$

because $$s \in S$$ (notation of Theorem 3.33) must be $$x$$, as far as I can see, because

$$\tilde{ \phi } (x_i) = s_i $$

becomes, in Theorem 3.97, $$\phi^*(x) = x$$

... but $$x \notin \mathbb{F}_p$$Can someone please clarify this for me ... and explain exactly how Theorem 3.33 applies to the prrof of Theorem 3.97

Peter

 

[Deveno]

I think you're right-Rotman must be quoting the wrong theorem.

The theorem I'm familiar with goes like this:

If $\phi: R \to S$ is a ring-homomorphism (of commutative rings with unity), then we have a ring-homomorphism:

$\phi_{\ast}: R[x] \to S[x]$ given by:

$\phi_{\ast}(a_0 + a_1x + \cdots + a_nx^n) = \phi(a_0) + \phi(a_1)x + \cdots + \phi(a_n)x^n$.

That said, he may thinking of this:

We know we have a ring-homomorphism $\Bbb Z \to \Bbb F_p$, and we know we have a ring-homomorphism:

$\Bbb F_p \to \Bbb F_p[x]$ (this is just inclusion). The composition of those yields the desired ring-homomorphism.

In any case, the important thing is, if we have a factorization in $\Bbb Z[x]$, reducing the coefficients of each factor mod $p$ yields a factorization in $\Bbb F_p[x]$. This is because the coefficients of polynomial sums and products are algebraic expressions of ring sums and products, which are preserved by the underlying ring-homomorphism $\Bbb Z \to \Bbb F_p$.

Note that by "reducing the coefficients mod $p$" we project a polynomial over a ring (the integers) to a polynomial over a finite field, in which it may be far easier to prove irreducibility, due to there being a finite number of possible factors.

 

[Math Amateur]

I think you're right-Rotman must be quoting the wrong theorem.

The theorem I'm familiar with goes like this:

If $\phi: R \to S$ is a ring-homomorphism (of commutative rings with unity), then we have a ring-homomorphism:

$\phi_{\ast}: R[x] \to S[x]$ given by:

$\phi_{\ast}(a_0 + a_1x + \cdots + a_nx^n) = \phi(a_0) + \phi(a_1)x + \cdots + \phi(a_n)x^n$.

That said, he may thinking of this:

We know we have a ring-homomorphism $\Bbb Z \to \Bbb F_p$, and we know we have a ring-homomorphism:

$\Bbb F_p \to \Bbb F_p[x]$ (this is just inclusion). The composition of those yields the desired ring-homomorphism.

In any case, the important thing is, if we have a factorization in $\Bbb Z[x]$, reducing the coefficients of each factor mod $p$ yields a factorization in $\Bbb F_p[x]$. This is because the coefficients of polynomial sums and products are algebraic expressions of ring sums and products, which are preserved by the underlying ring-homomorphism $\Bbb Z \to \Bbb F_p$.

Note that by "reducing the coefficients mod $p$" we project a polynomial over a ring (the integers) to a polynomial over a finite field, in which it may be far easier to prove irreducibility, due to there being a finite number of possible factors.

Thanks Deveno ... that clarifies issues significantly ...

Still thinking over all aspects of what you have said ... get main idea, though ...

Thanks again,

Peter