Rotman - Theorem 2.64 - Irreducibility

[Math Amateur]

I am reading Chapter 2: Commutative Rings in Joseph Rotman's book, Advanced Modern Algebra (Second Edition).

I am currently focussed on Theorem 2.64 [pages 115 - 116] concerning irreducibility.

I need help to the proof of this theorem.

Theorem 2.64 and its proof read as follows:https://www.physicsforums.com/attachments/2695
https://www.physicsforums.com/attachments/2696In the above text: Rotman writes:

" ... ... Suppose that \(\displaystyle f(x) \) factors in \(\displaystyle \mathbb{Z} [x] \); say \(\displaystyle f(x) = g(x)h(x) \), where \(\displaystyle deg(g) \lt deg(f) \) and \(\displaystyle deg(h) \lt deg(f) \). Now \(\displaystyle \overline{f}(x) = \overline{g}(x) \overline{h}(x) \) so that \(\displaystyle deg( \overline{f}) = deg( \overline{g}) + deg( \overline{h}). \) Now, \(\displaystyle \overline{f}(x) \) is monic because f(x) is and so \(\displaystyle deg( \overline{f}) = deg(f) \). Thus both \(\displaystyle deg( \overline{g}) \text{ and } deg( \overline{h}) \) have degrees less than \(\displaystyle deg( \overline{f}) \) ... ..."

My question is as follows:

How does (on what grounds) does Rotman reach the conclusion that both \(\displaystyle deg( \overline{g}) \text{ and } deg( \overline{h}) \) have degrees less than \(\displaystyle deg( \overline{f}) \)?

Indeed, what is preventing one of \(\displaystyle deg( \overline{g}) , deg( \overline{h}) \) from being equal to 1 and hence having degree zero and the other having degree n? (Having one of them having degree 0 and being equal to a unit other than 1 would, of course, contradict the monic nature of f, but one of them being equal to 1 seems an open possibility)

Hope someone can help clarify the above issue.

Peter

 

[Deveno]

A polynomial $f(x) \in F[x]$, for any field $F$ is said to be REDUCIBLE if:

$f(x) = g(x)h(x)$

where neither $g$ nor $h$ is a unit in $F[x]$ (a unit in $F[x]$ is, of course, a unit of $F$, that is to say: a non-zero constant polynomial).

If $f \neq 0$, then since $F[x]$ is an integral domain, if $f$ is reducible, neither $g$ nor $h$ can be 0, either.

Furthermore, if $f$ is monic, then without loss of generality we can find $g$ and $h$ monic as well, for if the leading coefficient of $g$ is $a$, then the leading coefficient of $h$ must be $1/a$, in which case we have:

$f(x) = [(1/a)g(x)][ah(x)]$ and both these factors are monic.

Now $\text{deg}(g),\text{deg}(h) > 0$, since these are non-zero non-units. This precludes the possibility that either one is the constant polynomial $1$.

In $\Bbb Z[x]$, the situation is considerably simpler: if such an $f$ factors into two non-unit polynomials OF LESSER DEGREE, then each factor must be AT LEAST degree 1. In other words: polynomials of degree one is "as low as we go"...the reason being, is that irreducibility is defined "only up to units".

We don't want factorizations like $f(x) = (-1)(-f(x))$ to "count", these are somewhat trivial. Again, this is along the same lines as not considering the integer 1 as "prime" for we can "factor it" all day long:

$1 = (-1)(-1)(-1)(-1)(1)(-1)(1)(-1)$, for example.

In a similar vein, in the ring $Z_5[x]$ we can factor the irreducible polynomial $x + 3$ as:

$x + 3 = (2)(3x + 4)$, but this really shouldn't count. Since $2 = 3^{-1}$ in $\Bbb Z_5$, the above factorization is actually:

$\dfrac{3x + 4}{3} = \dfrac{3x + 3}{3} + \dfrac{1}{3} = x + 1 + 2 = x + 3$.

IN GENERAL, for a commutative ring $R$, and an ideal $I$, if we have:

$f(x) = g(x)h(x)$ in $R[x]$, we certainly have:

$\varphi^{\ast}(f(x)) = \varphi^{\ast}(g(x))\varphi^{\ast}(h(x))$ in $R/I[x]$.

The underlying idea is this:

Often we want to establish a particular monic polynomial $f(x) \in \Bbb Z[x]$ is irreducible over $\Bbb Q$. Now Gauss' lemma tells us that if it factors in $\Bbb Q[x]$ (and, again "factors" means here "factors into two non-units", we want non-trivial factorizations), it factors in $\Bbb Z[x]$.

The ring-homomorphism $\varphi^{\ast}$ preserves the factorization, so if when we reduce mod $p$, we get an irreducible polynomial, we must not have had any possible factorization to begin with. The point to doing this, is that there often fewer cases to check in $\Bbb Z_p[x]$.

For example, take the polynomial:

$p(x) = x^3 - 2x^2 + 7x + 15 \in \Bbb Z[x]$.

The only thing immediately obvious is that any integral root is $\pm 1,3,5,15$. That's 8 possible roots to try. If we take this polynomial mod 2, we get:

$x^3 + x + [1]$, and we have just 2 roots to try:

$\overline{p}([0]) = [1]$
$\overline{p}([1]) = [1]$neither of which work, so we conclude that $p$ is irreducible over $\Bbb Q$.

In general, it's HARD to decide if a given monic integral polynomial is irreducible, especially if the degree is > 3. Even if we conclude it has no integer roots, it may still factor into integral quadratics, or cubic, etc. So irreducibilty criteria are invaluable (time-savers).

Indeed, the theorem you are looking at now is just a prelude to the important Eisenstein's criterion:

If $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \in \Bbb Z[x]$ is a polynomial such that:

$p|a_j$, for $j = 0,1,\dots,n-1$ but $p^2\not\mid a_0$, for some prime $p$, then $f(x)$ is irreducible over $\Bbb Q$.

To see how this works, suppose that on the contrary for such a polynomial:

$f(x) = g(x)h(x)$. Reducing mod $p$, we get:

$\overline{f}(x) = \overline{g}(x)\overline{h}(x)$.

Since $p$ divides all but the leading coefficient, we have $\overline{f}(x) = x^n$

Since $\Bbb Z_p[x]$ is a unique factorization domain, we must have $\overline{g}(x) = x^k$ for $0 < k < n$, and $\overline{h}(x) = x^{n-k}$.

But this implies the constant term of $g$ is divisible by $p$, and likewise for $h$, so that $p^2|a_0$, contradiction.

 

[Math Amateur]

A polynomial $f(x) \in F[x]$, for any field $F$ is said to be REDUCIBLE if:

$f(x) = g(x)h(x)$

where neither $g$ nor $h$ is a unit in $F[x]$ (a unit in $F[x]$ is, of course, a unit of $F$, that is to say: a non-zero constant polynomial).

If $f \neq 0$, then since $F[x]$ is an integral domain, if $f$ is reducible, neither $g$ nor $h$ can be 0, either.

Furthermore, if $f$ is monic, then without loss of generality we can find $g$ and $h$ monic as well, for if the leading coefficient of $g$ is $a$, then the leading coefficient of $h$ must be $1/a$, in which case we have:

$f(x) = [(1/a)g(x)][ah(x)]$ and both these factors are monic.

Now $\text{deg}(g),\text{deg}(h) > 0$, since these are non-zero non-units. This precludes the possibility that either one is the constant polynomial $1$.

In $\Bbb Z[x]$, the situation is considerably simpler: if such an $f$ factors into two non-unit polynomials OF LESSER DEGREE, then each factor must be AT LEAST degree 1. In other words: polynomials of degree one is "as low as we go"...the reason being, is that irreducibility is defined "only up to units".

We don't want factorizations like $f(x) = (-1)(-f(x))$ to "count", these are somewhat trivial. Again, this is along the same lines as not considering the integer 1 as "prime" for we can "factor it" all day long:

$1 = (-1)(-1)(-1)(-1)(1)(-1)(1)(-1)$, for example.

In a similar vein, in the ring $Z_5[x]$ we can factor the irreducible polynomial $x + 3$ as:

$x + 3 = (2)(3x + 4)$, but this really shouldn't count. Since $2 = 3^{-1}$ in $\Bbb Z_5$, the above factorization is actually:

$\dfrac{3x + 4}{3} = \dfrac{3x + 3}{3} + \dfrac{1}{3} = x + 1 + 2 = x + 3$.

IN GENERAL, for a commutative ring $R$, and an ideal $I$, if we have:

$f(x) = g(x)h(x)$ in $R[x]$, we certainly have:

$\varphi^{\ast}(f(x)) = \varphi^{\ast}(g(x))\varphi^{\ast}(h(x))$ in $R/I[x]$.

The underlying idea is this:

Often we want to establish a particular monic polynomial $f(x) \in \Bbb Z[x]$ is irreducible over $\Bbb Q$. Now Gauss' lemma tells us that if it factors in $\Bbb Q[x]$ (and, again "factors" means here "factors into two non-units", we want non-trivial factorizations), it factors in $\Bbb Z[x]$.

The ring-homomorphism $\varphi^{\ast}$ preserves the factorization, so if when we reduce mod $p$, we get an irreducible polynomial, we must not have had any possible factorization to begin with. The point to doing this, is that there often fewer cases to check in $\Bbb Z_p[x]$.

For example, take the polynomial:

$p(x) = x^3 - 2x^2 + 7x + 15 \in \Bbb Z[x]$.

The only thing immediately obvious is that any integral root is $\pm 1,3,5,15$. That's 8 possible roots to try. If we take this polynomial mod 2, we get:

$x^3 + x + [1]$, and we have just 2 roots to try:

$\overline{p}([0]) = [1]$
$\overline{p}([1]) = [1]$neither of which work, so we conclude that $p$ is irreducible over $\Bbb Q$.

In general, it's HARD to decide if a given monic integral polynomial is irreducible, especially if the degree is > 3. Even if we conclude it has no integer roots, it may still factor into integral quadratics, or cubic, etc. So irreducibilty criteria are invaluable (time-savers).

Indeed, the theorem you are looking at now is just a prelude to the important Eisenstein's criterion:

If $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \in \Bbb Z[x]$ is a polynomial such that:

$p|a_j$, for $j = 0,1,\dots,n-1$ but $p^2\not\mid a_0$, for some prime $p$, then $f(x)$ is irreducible over $\Bbb Q$.

To see how this works, suppose that on the contrary for such a polynomial:

$f(x) = g(x)h(x)$. Reducing mod $p$, we get:

$\overline{f}(x) = \overline{g}(x)\overline{h}(x)$.

Since $p$ divides all but the leading coefficient, we have $\overline{f}(x) = x^n$

Since $\Bbb Z_p[x]$ is a unique factorization domain, we must have $\overline{g}(x) = x^k$ for $0 < k < n$, and $\overline{h}(x) = x^{n-k}$.

But this implies the constant term of $g$ is divisible by $p$, and likewise for $h$, so that $p^2|a_0$, contradiction.

Thanks for the help Deveno ... ... just working through your post in detail now ...

Tanks again,

Peter

 

[Math Amateur]

Thanks for the help Deveno ... ... just working through your post in detail now ...

Tanks again,

Peter

Thanks again for the help, Deveno ... as I worked through your post and also checked on the Theorem 2.64 again, I became aware that the proof did not seem to explicitly make use of the fact that p is prime ... can you help?

My question is - why, exactly, does p have to be prime ... is it to ensure \(\displaystyle \mathbb{F}_p \) is a field, so that \(\displaystyle \mathbb{F}_p [x] \) is a domain? (Presumably we want \(\displaystyle \mathbb{F}_p [x] \) to be a domain - but why?)

Can you clarify this issue?

Peter

 

[Deveno]

It is a theorem that $\Bbb Z/n\Bbb Z$ is an integral domain if and only if $n$ is not composite (the trivial ring $\{0\}$ is "technically" an integral domain, but a somewhat boring one (this is the $n = 1$ case), and the integers themselves are an integral domain (this is the $n = 0$ case)).

Put another way: $\Bbb Z/n\Bbb Z$ is a non-trivial finite integral domain if and only if $n$ is prime.

To see this, suppose $n = ab$ with $1 < a,b < n$.

Then $[a] = [ab] = [n] = [0]$, so $[a]$ and $$ are zero divisors.

On the other hand if $n = p$, a prime, then $\text{gcd}(a,p) = 1$ for all $0 < a < p$, and thus $[a]$ is a unit, so there are no zero divisors.

Now if $n$ is composite, then $\Bbb Z_n[x]$ surely has zero divisors (for example, at least all the zero divisors of $\Bbb Z_n$, realized as constant polynomials). So $\Bbb Z_n[x]$ isn't even an integral domain, MUCH LESS an unique factorization domain. And if we don't have a UFD, we can't test for irreducibility by testing for primality (primality is easier to test for than irreducibility).

The presence of zero divisors in a ring is generally unhelpful to solving problems. For example, if $a$ is a zero divisor in a ring $R$, we can have $ab = ac$, with $b \neq c$. So even if we know $ab = x$, and we know what $a$ and $x$ are, we don't have a way to find out what $b$ is.