4
$\begingroup$

In text book "A course in abstract algebra" by author Khanna & bhambri" it is given that,

f(x) = 2( x^2) + 2 is irreducible polynomial over Z.

Because they used the definition "let R be an Integral domain with unity then a polynomial f(x) in R[x] of positive degree (i.e. deg ≥ 1) is said to be irreducible polynomial over R if it can not be expressed as product of two polynomials of positive degree. In other words, if whenever f(x) = g(x)• h(x) Then deg g = 0 or deg h = 0.

Here f(x) = 2(x^2) + 2 = 2(x^2 + 1) Clearly deg g = deg(2) = 0. So f(x) is irreducible polynomial over Z by above definition.

But in book "Contemporary abstract algebra" by "Joseph A. Gallian", they used the definition,

"Let D be an integral domain. A polynomial f(x) in D[x] that is neither a zero polynomial nor unit in D[x] is said to be irreducible over D if, whenever f(x) = g(x) • h(x) with, g(x) & h(x) from D[x], then g(x) OR h(x) is unit in D[x]."

Now here f(x) = 2(x^2) + 2 = 2( x^2 + 1)

Clearly neither 2 nor x^2 + 1 in Z[x] are unit. So that f(x) is reducible by definition of book by "Joseph A. Gallian".

So one book says, it is irreducible over Z but other says it is reducible over Z. Please suggest me which one I should prefer? .

$\endgroup$
5
  • $\begingroup$ Gallian's definition is more standard. It matches the general definition of irreducibility of an element in an arbitrary commutative ring with unity. $\endgroup$
    – quasi
    Mar 16 '17 at 3:24
  • $\begingroup$ Note that $2(x^2+1)$ is irreducible in $\mathbb{Q}[x]$. $\endgroup$
    – quasi
    Mar 16 '17 at 3:27
  • $\begingroup$ Yes sir, Book by Gallian is standard book and should follow its definition. But other that I had mention. It is also a well known book "of Delhi university". How that book is wrong? $\endgroup$ Mar 16 '17 at 3:29
  • $\begingroup$ Authors occasionally overlook details. In this case, the author conflated irreducibility in $\mathbb{Z}[x]$ with irreducibility in $\mathbb{Q}[x]$. Of course, you could also say it's "author's choice", but in this case, the choice is inconsistent with the standard meaning for general commutative rings with unity. $\endgroup$
    – quasi
    Mar 16 '17 at 3:32
  • $\begingroup$ Nevertheless, in any given course with a textbook, use the definitions specified in the text for the duration of the course, unless the teacher indicates otherwise. $\endgroup$
    – quasi
    Mar 16 '17 at 3:35
3
$\begingroup$

To dramatize the flaw in the definition given in the text by "Khanna & Bhambri" (K &B), consider the polynomials $$x,\;2x,\;3x,\;6x$$ By K&B's definition, the above polynomials are all irreducible in $\mathbb{Z}[x]$. Moreover, since none of them is a unit factor times one of the others, they would be regarded as distinct irreducibles (i.e., none is an "associate" of any of the others).

But then the polynomial $6x^2$ factors in $\mathbb{Z}[x]$ in two different ways

$$6x^2 = (2x)(3x)\qquad\text{and}\qquad 6x^2 = (x)(6x)$$

as a product of irreducible elements, thus breaking "unique factorization".

As I mentioned in my prior comment, it appears that the K&B text (perhaps accidentally) conflated irreducibility in $\mathbb{Z}[x]$ with irreducibility in $\mathbb{Q}[x]$.

I would regard it as an error, and use the standard definition instead (e.g., Gallian's definition), but check to make sure your teacher agrees.

In any case, good catch!

$\endgroup$
4
  • $\begingroup$ Just, I had asked about this to my teacher, they said Gallian definition is standard one but if we apply the Gallians definition then we saw, the polynomial f(x) = 2(x^2) + 2 = 2(x^2 + 1) is irreducible over Q (since 2 is unit in Q[x]) whereas it is reducible over Z(since neither 2 nor x^2 + 1 are units in Z[x] ). Hence this shows the polynomial is reducible over smaller field Z but not on bigger field Q, which is somehow looking absurd!! So they said go with bhambri definition. Now I am little bit confused? $\endgroup$ Mar 16 '17 at 12:48
  • $\begingroup$ Just, I had asked about this to my teacher, they said Gallian definition is standard one but if we apply the Gallians definition then we saw, the polynomial f(x) = 2(x^2) + 2 = 2(x^2 + 1) is irreducible over Q (since 2 is unit in Q[x]) whereas it is reducible over Z(since neither 2 nor x^2 + 1 are units in Z[x] ). Hence this shows the polynomial is reducible over smaller set Z but not on bigger set Q(though Z subset of Q) , which is somehow looking absurd!! So they said go with bhambri definition. Now I am little bit confused? $\endgroup$ Mar 16 '17 at 13:10
  • 1
    $\begingroup$ Well, I doubt that you'll find bhambri's definition used in any standard textbook on Abstract Algebra or Commutative Ring Theory, so you can regard that strange definition as temporary for your current course only. Hopefully, there won't be too many other non-standard definitions used in your course. $\endgroup$
    – quasi
    Mar 16 '17 at 19:56
  • 1
    $\begingroup$ @Akash: To illustrate the issue, let $A = \mathbb{Z}[1/2]$ (the smallest subring of the field $\mathbb{Q}$ containing the elements $1$ and $1/2$). Then $\mathbb{Z}$ is a subring of $A$, and $A$ is a subring of $\mathbb{Q}$. Applying the standard definitions of reducibility and irreducibility, $6$ is reducible in $\mathbb{Z}$, irreducible in $A$, and a unit in $\mathbb{Q}$ (hence neither reducible nor irreducible), so the phenomenon you describe is already present in this example. $\endgroup$
    – quasi
    Mar 16 '17 at 20:55
2
$\begingroup$

I think Gallian's definition is the more standard definition for the general definition of irreducible. One typically defines a non-zero non-unit $a\in R$ to be irreducible if $a=bc$ implies $(a)=(b)$ or $(a)=(c)$. Things are a bit more complicated if the ring has zero-divisors.

This definition means $2(x^2+1)$ is not irreducible since it factors into $2\cdot (x^2+1)$ and $2(x^2+1) \subsetneq (2)$ and $2(x^2+1) \subsetneq (x^2+1)$. You see though that if you are looking at polynomials over a field, the definitions coincide since any non-zero, degree $0$ polynomial would be a unit and $(a)=(\lambda a)$ for any unit $\lambda$ and any element $a \in R$.

I should elaborate on why this definition is the same as Gallian's in $D[X]$ where $D$ is a domain. If suppose $f$ is irreducible and $f=gh$. Then suppose WLOG, $(f)=(g)$ This means $g=fd$ for some polynomial $d$. But then $f=(fd)h$, cancellation (since $D[X]$ is a domain if and only if $D$ is) implies $dh=1$ so $h$ is unit. Clearly the converse holds. If $f=gh$ implies $g$ or $h$ is a unit, then $(f)=(h)$ or $(f)=(g)$, respectively.

$\endgroup$
2
  • $\begingroup$ Thanks all of you... In the first book that I had mentioned (of bhambri) they said.. as f(x) = 2 x^2 + 2 = 2(x^2 + 1) and deg 2 = 0 so that it is irreducible polynomial over Z but as 2 is not a unit hence, f(x) = 2 x^2 + 2 is "not" an irreducible element over Z. So this author(bhambri) used the term irreducible element and here they shown that "" every irreducible polynomial need not be an irreducible element"". So finally once again, now anyone of you tell me which one should I prefer...? $\endgroup$ Mar 16 '17 at 4:03
  • $\begingroup$ I am not following what you mean or are asking. Are you trying to distinguish between whether $2$ is irreducible over $\mathbb{Z}$ vs $\mathbb{Z}[X]$? If so, $2$ is definitely prime as an element of $\mathbb{Z}$ (since $\mathbb{Z}/(2)$ is field and thus a domain) and all primes are irreducible. This also proves that it is prime over $\mathbb{Z}[X]$ since $\mathbb{Z}[X]/(2) \cong \mathbb{Z}_2[X]$ which is a domain. $\endgroup$
    – CPM
    Mar 16 '17 at 4:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.