Intereting Posts

If $\lim_{n\to\infty}a_{n}=l$, Then prove that $\lim_{n\to\infty}\frac{a_{1}+a_2+\cdot..+a_n}{n}=l$
Spectral Measures: Square Root
Representation of a linear functional in vector space
What are some applications of elementary linear algebra outside of math?
To compute $\tan1-\tan3+\tan5-\cdots+\tan89$, $\tan1+\tan3+\tan5+\cdots+\tan89$
Find the volume of the set $S=\{x=(x_1,x_2,\cdots,x_n)\in \Bbb{R}^n:0\le x_1\le x_2\le \cdots \le x_n \le 1\}$
What is the value for $\lim_{x\to\infty} \frac{\sin x}{x}$?
Two definitions of Taylor polynomials
how to find the root of permutation
proving that $(n-1)^n>n^{n-1}$
An example of prime ideal $P$ in an integral domain such that $\bigcap_{n=1}^{\infty}P^n$ is not prime
Confusion about the definition of function
About the strictly convexity of log-sum-exp function
Does $x/yz$ mean $x/(yz)$ or $(x/y)z$?
Working with binomial coefficient $\sum_{k=0}^n (-1)^k \binom nk=0$

If a polynomial is irreducible in $R[x]$, where $R$ is a ring, 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.

- Why is quaternion algebra 4d and not 3d?
- Order of $\mathrm{GL}_n(\mathbb F_p)$ for $p$ prime
- Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?
- If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative.
- How to prove that the Frobenius endomorphism is surjective?
- Can someone explain the precise difference between of direct sum and direct product of groups?

- Direct limit of $\mathbb{Z}$-homomorphisms
- Is there a simple explanation why degree 5 polynomials (and up) are unsolvable?
- Find a basis of the $k$ vector space $k(x)$
- Question on Groups $G=\langle x,y|x^4=y^4=e,xyxy^{-1}=e\rangle$
- Binary operation (english) terminology
- Converse to Chinese Remainder Theorem
- Show that any prime ideal from such a ring is maximal.
- Abelianization of free group is the free abelian group
- Why free presentations?
- Show that $M$ is a Noetherian $A$-module.

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 $\mathbb{Q}[x]$.

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.

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 $f(2)=0$.

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.

- Non-existence of a bijective analytic function between annulus and punctured disk
- Definite Integral $\int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(3+x)}}dx$
- Find an algorithm to evaluate unknown polynomial of degree $n$ given its values for $x=0,x=1, x=2,\ldots,x=n$
- Use implicit function theorem to show $O(n)$ is a manifold
- Constructing self-complementary regular graphs
- Rubik's Cube Not a Group?
- Motivation for different mathematics foundations
- $2^{nd}$ order ODE $4x(1-x)y''-y=0$ with $y'(0)=1$ at $x=0$
- A certain problem concerning a Hilbert class field
- Proving that any rational number can be represented as the sum of the cubes of three rational numbers
- Alternative proof that Harmonic sum is not an integer
- Inequality:$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$
- Did Galois show $5^\sqrt{2}$ can't solve a high-order integer polynomial?
- Combinatorics question about english letters (with consonants and vowels)
- Does $\sum_{n=1}^\infty \frac{1}{n! \sin(n)}$ diverge or converge?