If a polynomial is irreducible in $R[x]$, where $R$ is a ring, it means that it does not have a root in $R$, right?

For example, to say that a polynomial $f(x)\in\mathbb Z[x]$ is irreducible in $\mathbb Q[x]$ is equivalent to say that $f(x)$ does not have any rational root. I just want to make sure.


5 Answers 5


An element $a$ of any ring (including polynomial rings) is reducible if and only if there exist elements $b$ and $c$ such that

  • $a = bc$
  • $b$ is not invertible
  • $c$ is not invertible

In the special case of polynomial rings over fields, an element (i.e. a polynomial) $f$ is reducible if and only if there exist non-constant polynomials $g$ and $h$ such that $f = gh$. This is because the non-zero constants are precisely the invertible elements.

The condition of being irreducible if it doesn't have any roots is false. Consider, for example, the polynomial

$$ x^4 + 4 x^2 + 3 = (x^2 + 1)(x^2 + 3) \in \mathbb{R}[x] $$

When the coefficient ring is not a field, though, some coefficients are not invertible. The polynomial

$$ 2x \in \mathbb{Z}[x]$$

is reducible, because it is the product of $2$ and of $x$, both of which are not invertible. However, $2x \in \mathbb{Q}[x]$ is irreducible; the key difference is in this latter case, $2$ is invertible. Also, note that $2x$ has a rational root, despite being irreducible in $\mathbb{Q}[x]$.

  • 2
    $\begingroup$ He is not asking whether not having a real root implies irreducibility. He is asking about the other way around. $\endgroup$
    – sxd
    Commented Oct 27, 2012 at 19:30
  • 5
    $\begingroup$ He uses "equivalent" in his example question. $\endgroup$
    – user14972
    Commented Oct 27, 2012 at 19:31
  • 1
    $\begingroup$ then his question is inconsistent :P $\endgroup$
    – sxd
    Commented Oct 27, 2012 at 19:51
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    $\begingroup$ This is because all non-zero constants are invertible. It is not because all non-zero constants are invertible but they are the only invertible elements of the field. $\endgroup$
    – synack
    Commented Oct 27, 2012 at 21:56
  • $\begingroup$ @Kits: Good point. My comment was meant to focus on one direction, but it's easy enough to correct it to apply on both directions, so I shall do so! $\endgroup$
    – user14972
    Commented Oct 27, 2012 at 22:11

What you're asking is almost true. An irreducible polynomial has a root if and only if it is linear. Proof:

Let $k$ be an integral domain. Assume that $f\in k[x]$ is irreducible, i.e. whenever $f=gh$, then either $g$ or $h$ is a unit. Assume that $a\in k$ is a root of $f$, i.e. $f(a)=0$. We perform polynomial division of $f$ by $(x-a)$, yielding $f=(x-a)g + r$ with $\deg(r)<\deg(x-a)=1$, meaning $r\in k$. Since $0=f(a)=r(a)$, it follows that $r=0$ and hence, $f=(x-a)g$ with $g\in k[x]$. But since $f$ is irreducible, this means that $g$ is a unit, i.e. $f$ is a linear polynomial.

  • $\begingroup$ To perform polynomial division, don't we need that $k[x]$ is an Euclidean domain? For example, would this still work in $\mathbb{Z}[x]$? $\endgroup$
    – Anakhand
    Commented Sep 29, 2023 at 20:12
  • 1
    $\begingroup$ Dear @Anakhand; You do not need to be in a Euclidean domain to perform polynomial division by a monic polynomial like $x - a$. More generally, you can perform polynomial division by $p$ if the leading coefficient of $p$ is a unit over any integral domain. $\endgroup$ Commented Oct 2, 2023 at 13:15
  • $\begingroup$ So yes; this would work over $\mathbb{Z}[x]$. Problems arise when you try to divide something like $x-1$ by $2x-1$, which would not have the desired degree reduction property for the remainder. However, there is no problem dividing $2x-1$ by $x-1$, and I hope that looking at this example illustrates how you can really divide any polynomial over $\mathbb{Z}[x]$ by $x-a$ for $a\in\mathbb{Z}$. $\endgroup$ Commented Oct 2, 2023 at 13:18

I know there's an accepted answer here, but I just wanted to add in something to clarify a couple answers for newcomers:

If $F$ is a field, $f(x)\in F[x]$ is reducible if and only if $f(x)$ has a zero in $F$, but this is only always true for polynomials of degree 2 and 3.

Mark Bennet gives a decent counterexample to the generalized claim, and note that the polynomial he uses is degree 4.

However, things are a bit different when you're working in $Z_n$ ($Z/nZ$). You can check for reducibility by testing if $f(n)=0$ for $n\in[0,n-1]$.

For example, $f(x)=x^3+1\in Z_9[x]$ is reducible over $Z_9$ because $f(2)=0$.

  • $\begingroup$ is second result you gave for $Z_{n}$ true for any order ? $\endgroup$
    – ogirkar
    Commented Apr 17, 2019 at 14:52

To look at things another way $$x^4+5x^2+4=(x^2+1)(x^2+4)$$

is reducible over $\mathbb R$, but does not have any real roots.


Suppose $p(a)=0$ when $p(x)$ is a polynomial over a (commutative) ring. Then $p(x)=p(x)-p(a)$ and the fact that $x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+\dots+a^{n-1})$ for all $n$ means that $(x-a)$ is a linear factor of $p(x)$ - so if the polynomial has degree greater than 1 it is reducible.


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