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The definition claims that a polynomial in a field of positive degree is a reducible polynomial when it can be written as the product of $2$ polynomials in the field with positive degrees. Other wise it is irreducible.

So if a polynomial $f(x)$ can be written as the product of say $41(x^2 + x)$, is that considered not reducible because $41$ is really $41x^0$, and $0$ isn't technically positive, but by the definition of a polynomial in a field it is a polynomial if $a_n$ isn't $0$ for the highest degree $n$ where $n \geq 0$.

So would the example of the polynomial example I gave be reducible or irreducible?

L. F.
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ming
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  • I recommend writing zero in dollars because the zero character is like the letter o in the default font which makes reading difficult. Compare 0 and $0$ if you use the default font. – L. F. Nov 28 '18 at 10:59

2 Answers2

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This particular example would be reducible, but not because of the reason you give -- it would be because you could write it as $41x(x+1)$. Note that $0$ is not positive, so the factorization you give is not a product of two polynomials with positive degree ($41$ is a polynomial of degree $0$).

platty
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    So if it was $41(x + 1)$ it would be irreducible right? Because 41 cannot be written with a degree of anything greater than 0, and 0 is not positive? – ming Nov 28 '18 at 06:23
  • Correct, $41x + 41$ is irreducible (assuming you are working over $\mathbb{Z}[x]$). – platty Nov 28 '18 at 06:25
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    It is reducible in Z[x] as 41 is not a unit. – Hans Nov 28 '18 at 07:33
  • I think the definition of irreducible as a polynomial requires that both factors be non-constant; I'm not sure why I added that stipulation. Otherwise you run into uniqueness issues with something like $4x+4$. – platty Nov 28 '18 at 07:37
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Definition : Given an integral domain $R$, a non-zero non-unit element $r \in R$ is said to be irreducible if it cannot be written as a product of non-units i.e. whenever it is written as a product of two elements, at least one of them is a unit in $R$.

For polynomials, this becomes :

Definition : Given an integral domain $R$, the ring $R[X]$ also is an integral domain, and $f$ is an irreducible polynomial over $R$ if it is an irreducible element of $R[X]$.

So, it is that simple. Let us take some examples to clarify.

  • The polynomial $f(x) = x$ is irreducible over any ring, since if $a(x)b(x) =x$, then WLOG $a$ must be a constant polynomial with the constant being a unit(use the rules for multiplication of polynomials), so $a$ is a unit in $R[X]$, hence $x$ is irreducible.

  • The polynomial $f(x) = 2x+2$ is irreducible over $\mathbb R[X]$. This is because if $a(x)b(x)$ divides $2(x+1)$ then at least one of $a(x)$ or $b(x)$ is a constant polynomial, but every constant polynomial is a unit in $\mathbb R$. However, this polynomial is reducible over $\mathbb Z[X]$, since here, $2(x+1)$ counts as a non-unit factorization, because $2$ is not a unit.

Therefore, reducibility depends on "over which ring/field"? For example, $41 = 41x^0$ is a constant polynomial, but it isn't a unit in $\mathbb Z[X]$, while it is one in $\mathbb R[X]$. So, a polynomial like $41(x+1)$ is irreducible over the latter but not over the former.

While working over a field, it turns out that the set of units of $F[X]$ is equal to the non-zero constant polynomials. Therefore, any polynomial is irreducible in $F[X]$ if and only if it can be written as the product of two non-constant polynomials. Things will change for a ring which is not a field, since some non-constant polynomials may possibly not be units.

Also, note that $41(x^2+x)$ is reducible in every field where it is non-zero, since it can be written as $(41 x) \times (x+1)$ which is the product of two non-constant polynomials, which are always non-units by the fact that the degree is multiplicative.

The reason why "where it is non-zero" is mentioned, is because in characteristic $41$(or, over fields of characteristic $41$ such as $\mathbb Z \over 41\mathbb Z$), this polynomial actually reduces to $0$, for which the notion is irreducibility is not defined above. For fields of any other characteristic, the polynomial is non-zero(and non-unit) so we can discuss irreducibility. Much thanks to the comment below for pointing this out. Also, something similar applies to other polynomials discussed previously such as $2x+2$ (in characteristic $2$, this polynomial is identically zero) and $41(x+1)$. However, $x$ remains non-zero over every field, so the argument remains.

Sarvesh Ravichandran Iyer
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    There is a small subtlety in your last paragraph over fields of characteristic $41$ that would be worth mentioning. – Eric Towers Nov 28 '18 at 13:47
  • @EricTowers It is very important, thank you for mentioning it. I have edited the answer. – Sarvesh Ravichandran Iyer Nov 28 '18 at 13:58
  • So polynomial can be reducible over smaller set $\mathbb{Z}$ even Though it is irreducible over bigger set $\mathbb{Q}$. So polynomial which can not be factored in $\mathbb{Q}$ can be factored in $\mathbb{Z}$. Isn't it looks absurd. – Akash Patalwanshi Sep 15 '21 at 05:04
  • @AkashPatalwanshi It sounds funny, and I sort of see why : if $p$ is reducible over the integers, and you write a polynomial $p(x) = q(x)r(x)$ where $q,r$ have integer coefficients, then of course $q$ and $r$ also have rational coefficients, so $p$ should be reducible over the rationals, right? Unfortunately, the notion of a unit implies that even if $q,r$ have rational coefficients, they could potentially become units because $\mathbb Q$ admits more units than $\mathbb Z$. For example, $2$ is invertible in $\mathbb Q$ but not in $\mathbb Z$. It barely disturbs life, to be fair... – Sarvesh Ravichandran Iyer Sep 15 '21 at 05:35
  • ... The reason being that if $q,r$ are non-constant polynomials, then they cannot be units. So the only problem that can occur when one talks about changing the base of the ring, comes with respect to the units of the underlying rings, and therefore only constant polynomials can cause any headache. If constant polynomials are absent, then the expressions may be handled as if the base ring is the same. – Sarvesh Ravichandran Iyer Sep 15 '21 at 05:37
  • Sorry to comment on this old post. I have a question about the definition of irreducible elements in an integral domain. Why do we only define irreducibility for non-unit elements? What problem will happen if we try to apply the same definition to unit elements? Thanks. – Sam Wong Dec 20 '22 at 07:16
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    @SamWong Thank you for your question. What will happen is that the "usefulness" of irreducibility will break down : for example, see here and this one as well. Basically, we need to stop at units because if we involved units as well, then pretty much anything can be factorized as we please. For example, if you look at factoring $x+1$ over the rational numbers, then any of $n(\frac{x}{n}+\frac 1n)$ would be acceptable factorizations, and uniqueness of factorization would be lost. – Sarvesh Ravichandran Iyer Dec 20 '22 at 16:28
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    To me at least, the existence of factorizations is not an issue. However, if one chooses to factor units, then uniqueness absolutely falls apart. There needs to be a point where one can say that a particular ring element is "completely factored" and based on what we know from the integers (where we don't write $1 \cdot 2 \cdot 3$ as tbe alternate factorization $-1 \cdot -1 \cdot 2 \cdot 3$, for example) units are the "right" place to stop. Of course, feel free to continue clarifying things, it's been some time since I've done algebra. – Sarvesh Ravichandran Iyer Dec 20 '22 at 16:31