Is the polynomial $ x^4+3x^2+2 \ $ is irreducible over $ \mathbb{Q}[x] \ $ ?



Since both of the factors $x^2+1$ and $x^2+2$ have no roots in $\mathbb Q[x]$, so the polynomial is irreducible in $ \mathbb{Q}[x]. $

But if I do the modulo $2$ test in $\mathbb{Z}_2,$ then the polynomial reduces to $ x^4+x^2 \ =x^2(x^2+1),$ which is reducible in $\mathbb Z_2[x].$

Since the polynomial is a primitive polynomial, it is reducible in $\mathbb Q[x]$ also.

I am little confused , which one is coorect ?

  • 1
    $\begingroup$ Hint: What is the definition of "irreducible"? $\endgroup$ – Daniel Robert-Nicoud Sep 17 '17 at 15:47
  • $\begingroup$ $ f(x) \ $ is irreducible if $ f(x)=p(x)q(x) \ $ , their one of $ p(x) \ or \ q(x) \ $ must be unit . $\endgroup$ – M. A. SARKAR Sep 17 '17 at 15:50
  • 1
    $\begingroup$ Irreducible is not the same as "has no roots." And it is unclear what you mean by "...has no roots in $\mathbb Q[x].$" Do you mean "... has no roots in $\mathbb Q$?" $\endgroup$ – Thomas Andrews Sep 17 '17 at 15:50
  • $\begingroup$ which one of my works is correct ? $\endgroup$ – M. A. SARKAR Sep 17 '17 at 15:55

You factored it; you reduced it. So it's not irreducible. Being reducible does not mean that it factors "all the way" into linear factors. Any potential factoring (aside from factoring out a unit scalar) makes it reducible.

  • $\begingroup$ so what would be the answer? The second one of my work is correct . $\endgroup$ – M. A. SARKAR Sep 17 '17 at 15:56
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    $\begingroup$ Your sentence "Since both of..., so the polynomial is irreducible..." is not correct. Because you are saying that being irreducible depends on having roots. You have answered the question as soon as you factored it as $(x^2+1)(x^2+2)$. You can then immediately conclude the polynomial is reducible. And it is then unnecessary to do anything modulo $2$. $\endgroup$ – alex.jordan Sep 17 '17 at 16:28
  • $\begingroup$ I am confused about the two ways . I am not sure which should be correct . $\endgroup$ – M. A. SARKAR Sep 17 '17 at 16:36
  • $\begingroup$ Your "way" that says "Since..." is incorrect. You are missing the point that there is a more immediate "way" to answer the question. $\endgroup$ – alex.jordan Sep 17 '17 at 19:55
  • $\begingroup$ Your "way" that mentions reducing it mod $2$ is also incorrect. Just because something is reducible over $\mathbb{Z}_2$ doesn't mean it reduces over $\mathbb{Q}$. Consider $x^2+1$. $\endgroup$ – alex.jordan Sep 17 '17 at 19:59

you make two major mistakes:

1) to be irreducible, and not have a root are not the same thing.

$P\in\mathbb{Q}[X] $ irreducible $\Rightarrow$ $P$ has no root in $\mathbb{Q}$ is true, but the converse implication is false in general (except for degree $2$ and $3$). $P=(x^2+1)(x^2+2)$ has no root but is reducible (see below).

To be irreducible for a polynomial $P$ means by definition that $P$ is not constant, and cannot be factored into the product of two non constant polynomials.

2) The mod $p$ test says :If a primitive polynomial whose leading term is not divisible by $p$ is irreducible mod $p$, then it is irreducible.

The converse implication is not true: $x^2+1$ is reducible mod $2$ (because it is $(x+1)^2$ mod $2$, but it is irreducible (exercise).

To conclude, i am afraid that none of your reasonings is correct, sorry...

  • $\begingroup$ according to mod 2 test , the polynomial $ x^4+3x^2+2 \ $ reduces to $ x^4+x^2 \ =x^2(x^2+1)=x^2 (x+1)^2=x^2(x+1)(x+1)$ , Since each of $ x^2, \ x+1 , \ $ are non-unit in $ Z_2 \ $ , the polynomial is irreducible $\endgroup$ – M. A. SARKAR Sep 17 '17 at 16:11

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