Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?

$F[x]/\langle p(x)\rangle $ does not have to be a field. Could you name an example, preferably one for a field another for a non-field just so I can see the difference?

It was suggested by Arturo that I should re-edit this question since it was not clear enough: Let F be a field, and say $a(x)∈F[x]$ be irreducible. If $p(x)$ is not constant and not equal to an associate of $a(x)$, and we let $R=F[x]/(p(x))$, and identify F with its image in R, can $a(y)$ be reducible in $R[y]$?

share|cite|improve this question
I think there may be a problem with the question you've posed. I assume you mean $\mathbb{F}$ is a field. Correct? If so, $\mathbb{F}[x]$ is a principal ideal domain and as such every non-trivial quotient has the property that every non-zero element is either a zero divisor or a unit. Thus if a quotient lacks zero divisors, it's a field. We usually only speak of irreducibles in the context of an integral domain. So in your context speaking of a reducible/irreducible element only makes sense if the quotient is also a field. In this case, there are no irreducibles (only units and zero). – Bill Cook Oct 22 '11 at 2:08
Oops! I guess I read the title of your question and not the contents. My comment addresses the issue with your title. I'll give you an example addressing your question below. – Bill Cook Oct 22 '11 at 2:14
I understand your first comment. It was helpful. – crazy student Oct 22 '11 at 2:44
Please make sure the question you ask is actually in the post, not just in the title. – Arturo Magidin Oct 22 '11 at 3:27
@Bill: Actually, it is perfectly sensible to talk about irreducible and prime elements in the context of any (commutative) ring; on the other hand, talking about reducible elements (and reducible polynomials) is a rather different kettle of fish. – Arturo Magidin Oct 22 '11 at 4:10
up vote 1 down vote accepted

First, let's be clear:

If $R$ is a commutative ring with $1$, then an element $a\in R$ is irreducible if and only if $a$ is not a unit, $a$ is not a zero divisor, and whenever $a=bc$ in $R$, either $b$ is a unit (invertible) or $c$ is a unit. Added. Elements are not usually called "reducible", though. Elements that are not irreducible are "not irreducible".

In the particular case of $R=F[x]$ with $F$ a field, this agrees with the "usual" notion of "irreducible polynomial", since the units are precisely the nonzero constant polynomials.

Now, if $F$ is a field, then $F[x]$ is a unique factorization domain (even more, a Euclidean domain). If $a(x)$ is irreducible, then it is a prime in $F[x]$. So for any $p(x)\in F[x]$ there are only three possibilities: either $a(x)$ divides $p(x)$, or $p(x)=0$, or else $a(x)$ and $p(x)$ are relatively prime, and in particular, there exist polynomials $r(x)$ and $s(x)$ such that $$a(x)r(x) + p(x)s(x) = 1.$$

Now, suppose that $p(x)$ is a polynomial. The only way for $a(x)$ to be a multiple of $p(x)$ is for $a(x)$ to be of the form $a(x)=\lambda p(x)$ for some nonzero constant $\lambda\in F$. Assume that $a(x)$ is not a multiple of $p(x)$.

If $p(x)\neq 0$ and $a(x)$ divides $p(x)$, then $p(x) = a(x)q(x)$ for some $q(x)\notin F$. We cannot have $p(x)$ dividing $q(x)$, so then $\overline{q(x)}$ is not zero in $F[x]/\langle p(x)\rangle$. Neither is $\overline{a(x)}$. Yet $$\overline{a(x)}\times\overline{q(x)} = \overline{a(x)q(x)} = \overline{p(x)} = 0.$$ So $\overline{a(x)}$ is not irreducible, because it is a zero divisor.

If $a(x)$ does not divide $p(x)$, then we can write $1 = a(x)r(x)+p(x)s(x)$ for some polynomials $r(x)$ and $s(x)$. But then in $F[x]/\langle p(x)\rangle$ we have: $$1 = \overline{1} = \overline{a(x)r(x) + p(x)s(x)} = \overline{a(x)}\overline{r(x)} + \overline{p(x)}\overline{s(x)} = \overline{a(x)}\overline{r(x)},$$ so $\overline{a(x)}$ is not irreducible because it is a unit.

Finally, if $p(x)=0$, then $F[x]/\langle p(x)\rangle$ is essentially just $F[x]$ itself again, so $a(x)$ is irreducible in the image because it's just $a(x)$ in $F[x]$.

So, to answer the question that was posed, if $p(x)\neq 0$ then $a(x)$ will not be irreducible in $F[x]/\langle p(x)\rangle$ no matter what.

That said...

But your use of the word "reducible" suggests that you are actually wondering about something else: whether "as a polynomial" it can be written as a product of two nonconstant polynomials, or perhaps whether "it has a root" when considered in $F[x]/\langle p(x)\rangle$.

In other words: say $p(x)$ is a nonconstant polynomial in $F[x]$. Then $R=F[x]/\langle p(x)\rangle$ is a ring that contains a copy of $F$. We can then consider the polynomial $a(x)$ as having coefficients in $R$ and ask whether it is reducible or irreducible over $R$. However, using $x$ here is confusing because $x$ is already playing a role in the definition of $R$, so it's better to switch to a different letter in order to talk about "polynomials with coefficients in $R$". So let's use $y$ to denote polynomials with coefficients in $R$, and so we will be asking whether $a(y)$ is irreducible over $R$ or reducible over $R$.

If $p(x)$ is irreducible over $F$, then $F[x]/\langle p(x)\rangle$ is a field that contains $F$, and you can consider $a(y)$ as a polynomial in $(F[x]/\langle p(x)\rangle[y]=R[y]$. There, it may be that $a(y)$ is irreducible or that it is reducible.

For an example where it is reducible, take $a(x) = x^4 + 1$ in $\mathbb{Q}[x]$, and $p(x) = x^2+1$, both irreducible. Since $\mathbb{Q}[x]/\langle p(x)\rangle = \mathbb{Q}(i)$, the rational complex numbers, in $\mathbb{Q}(i)[y]$ we have $a(y) = y^4+1 = (y^2+i)(y^2-i)$, reducible.

For an example where it is irreducible, take $a(x) = x^2 - 2$ and $p(x) = x^2-3$ in $\mathbb{Q}[x]$. Then $\mathbb{Q}[x]/\langle p(x)\rangle$ is just $\mathbb{Q}(\sqrt{3})$, and $a(y)$ has no roots in $\mathbb{Q}(\sqrt{3})$; being degree $2$, it is irreducible in $\mathbb{Q}(\sqrt{3})[y]$.

If $p(x)$ is reducible, then you end up with a ring that is not a field and has zero divisors. In any case, as a consequence of the Chinese Remainder Theorem, if we factor $p(x)$ into irreducibles, $$p(x) = q_1(x)^{\alpha_1}\cdots q_m(x)^{\alpha_m},$$ with $q_i$ and $q_j$ coprime for $i\neq j$, $\alpha_i\gt 0$, then $$R=F[x]/\langle p(x)\rangle \cong \frac{F[x]}{\langle q_1(x)^{\alpha_1}\rangle}\times\cdots\times \frac{F[x]}{\langle q_m(x)^{\alpha_m}\rangle},$$ and this $a(y)$ may or may not be irreducible over this ring.

For an example where it is reducible, take the example above in which $a(y)$ was reducible, but replace $p(x)$ with $(x^4+1)(x^2+2)$ or some other reducible polynomial that has $x^4+1$ as a factor. Then $a(y)$ will have a root in the quotient, because it will have a root in one of the factors. Actually, since $F$ is embedded into the quotient "diagonally", you need two irreducible polynomials over which $a(x)$ was reducible, or a power of one over which $a(x)$ was reducible. For an example where it is irreducible, do the same but now take the product of two polynomials over which $a(y)$ was irreducible. you can take $F=\mathbb{Q}$, $a(x) = x^2-2$, $p(x) = (x^2-3)(x^2-5)$. Then $F[x]/\langle p(x)\rangle \cong \mathbb{Q}(\sqrt{3})\times\mathbb{Q}(\sqrt{5})$, and that ring does not have a square root of $2$ either, so $a(y)$ is still irreducible.

share|cite|improve this answer
I have a question about your a(x) = x^4 + 1 and p(x) = x^2 + 1. Aren't we modding out by p(x) so x^2 is really congruent to -1 (mod p(x))? So x^4 + 1 would be (x^2)(x^2) + 1 = 1 + 1 = 2? What's the y supposed to be, I'm kind of confused about that notation. – crazy student Oct 22 '11 at 4:37
@crazystudent: The problem is that your entire question is confused. Are you talking about "irreducibility of polynomials", or "irreducibility of elements"? If you are talking about elements, I answered the question in its entirety above the line. Since $x^4+1$ and $x^2+1$ are relatively prime, then $x^4+1$ becomes a unit in the quotient, so it cannot be irreducible (or "reducible", either). (cont) – Arturo Magidin Oct 22 '11 at 4:41
@crazystudent: If you are talking about polynomials, then you should not use $a(x)$ for the polynomial in the quotient, because $x$ represents something else. You would really be asking whether the polynomial $y^4+1$ is irreducible over $\mathbb{Q}[x]/(p(x))=\mathbb{Q}[i]$ or whether it is reducible. But if you really want to talk about irreducibe/reducible elements, then (i) You need to be clear; (ii) you need to distinguish the element of $F[x]$ with its image in the quotient; (iii) you need to clarify your notation, which is confused right now. – Arturo Magidin Oct 22 '11 at 4:43
@crazystudent: Then what you wrote is your question is nonsense. First, the correct preposition would be "irreducible over", not "irreducible in". Second, $a(x)$ is not a polynomial over $F[x]/\langle p(x)\rangle$. That's why I switched letters. If you let $R=F[x]/\langle p(x)\rangle$, then you can consider polynomials with coefficients in $R$. But using $x$ for the variable is very bad form and potentially confusing, because $x$ already is playing a different role. So instead I called the variable $y$, and talk about polynomials in $R[y]$ being irreducible. – Arturo Magidin Oct 22 '11 at 4:50
@crazystudent: Here's the phrasing you seem to be going for: I suggest editing your question and adding this below your current phrasing. "Let $F$ be a field, and say $a(x)\in F[x]$ be irreducible. If $p(x)$ is not constant and not equal to $a(x)$, and we let $R=F[x]/(p(x))$, and identify $F$ with its image in $R$, can $a(y)$ be reducible in $R[y]$?" Note that we don't look at the image of $a(x)$ in $R$, we are looking at the same polynomial, but now think of it as having coefficients in $R$. – Arturo Magidin Oct 22 '11 at 4:53

If you quotient by a reducible polynomial, you'll end up with zero divisors.

Consider $\mathbb{Q}[x]/(x^2-1)$. I'll refer to cosets "$f(x)+(x^2+1)$" just by their representatives "$f(x)$". Notice that $x+1,x-1$ are non-zero in this quotient, but $(x+1)(x-1)=x^2-1=0$ (mod $x^2-1$). So we've got zero divisors.

If you want a better feel for what this ring "really" looks like. It's not hard to show that $\mathbb{Q}[x]/(x^2-1) \cong \mathbb{Q}[x]/(x-1) \times \mathbb{Q}[x]/(x+1) \cong \mathbb{Q} \times \mathbb{Q}$. [Use the map $f(x) \mapsto (f(x)+(x-1),f(x)+(x+1))$ it's not hard to show this map is a homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[x]/(x-1) \times \mathbb{Q}[x]/(x+1)$ has kernel $(x^2-1)$ and using Chinese remaindering you can show it's onto. Then the result follows form the 1st isomorphism theorem.]

share|cite|improve this answer
Small typo on the second line :) – Alexei Averchenko Oct 22 '11 at 2:31
Chinese remaindering? I don't follow you when you say that Bill. I'm only taking an introductory abstract algebra course – crazy student Oct 22 '11 at 2:37
I just want a real example like functions a(x), p(x) and any ring F[x] – crazy student Oct 22 '11 at 2:40

I will address what I believe you are trying to ask for in the first scenario.

Let $F = \mathbb{Q}$, $a(x) = x^4-5$, $p(x) = x^2-5$. Note that both $a(x)$ and $p(x)$ are both irreducible by Eisenstein criterion. But $F[x]/(p(x)) = F[\sqrt{5}] = F(\sqrt{5})$, and over the polynomial ring $F(\sqrt{5})[x]$ we can factor $a(x) = (x^2-\sqrt{5})(x^2+\sqrt{5})$.

share|cite|improve this answer
Maybe I didn't get the memo here but it seems like everyone is saying x^4 - 5= (x^2 - sqrt(5))(x^2 + sqrt(5)) without mentioning that x^2 = 5 (mod p(x)). That is the only part that doesn't make sense to me. It's like everyone is ignoring that fact. – crazy student Oct 22 '11 at 4:45
@crazystudent: You need to make up your mind. What you are doing is viewing $a(x)$ as an element of the quotient, not as a polynomial. If you want to talk about irreducibility of polynomials, as you claim, then you do not view $a$ as an element of the quotient, you view it as a polynomial with coefficients in the quotient. That's why what you are writing is nonsensical: the $x$ in $a$ is an indeterminate, not an element of the quotient. Nobody is "ignoring that fact", we are trying to make sense of your confusion. – Arturo Magidin Oct 22 '11 at 5:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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