Intereting Posts

Solve $V_1+V_2+\cdots+V_k=A, V_1^2+V_2^2+\cdots+V_k^2=B$ in positive integers
Separable and non-separable function
Great books on all different types of integration techniques
Multivariable limit with logarithm
Galois Group of $X^4+X^3+1$ over $\mathbb{Q}$
Dirichlet's Divisor Problem
Prove there is no simple group of order $729$
A conceptual understanding of transmutations (and bosonizations) of (braided) Hopf algebras
Complex numbers and Roots of unity
The topological boundary of the closure of a subset is contained in the boundary of the set
Is there a classification of all finite indecomposable p-groups?
A polynomial that is zero on an open set
What does it mean to induce a topology?
A pair of continued fractions that are algebraic numbers and related to $a^2+b^2=c^m$
To Find The Exponential Of a Matrix

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.

- Factoring morphisms in abelian categories
- The total ring of fractions of a reduced Noetherian ring is a direct product of fields
- The fix points of the Möbius transformations are the eigenspace of a certain matrix.
- Matrices which commute with all the matrices commuting with a given matrix
- Element of order 4 in PSL(2,7)
- What does it mean to be an irreducible polynomial over a field? (need clarification on the definition)

- Is this an equivalent statement to the Fundamental Theorem of Algebra?
- Finding a primitive root modulo $11^2$
- Factoring morphisms in abelian categories
- Irreducibles are prime in a UFD
- Abstract algebra book recommendations for beginners.
- Symmetric polynomials and the Newton identities
- Reference request for ordered groups
- $G \cong G \times H$ does not imply $H$ is trivial.
- If $\lvert\operatorname{Hom}(H,G_1)\rvert = \lvert\operatorname{Hom}(H,G_2)\rvert$ for any $H$ then $G_1 \cong G_2$
- A morphism of free modules that is an isomorphism upon dualization.

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.

- Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.
- What are some interpretations of Von Neumann's quote?
- Number of subgroups of prime order
- $\ker \phi = (a_1, …, a_n)$ for a ring homomorphism $\phi: R \to R$
- Show that $504 \mid (n^9 − n^3 )$ for any integer $n$
- How to conceptualize unintuitive topology?
- Number of binary numbers with two consecutive zeros
- Counting distinguishable ways of painting a cube with 4 different colors (each used at least once)
- Show $\sum_{k=1}^{\infty}\left(\frac{1+\sin(k)}{2}\right)^k$ diverges
- Intuition — $c\mid a$ and $c\mid b$ if and only if $c\mid \gcd(a,b)$.
- Set Theory- Generalized form of distributivity of unions over intersections
- Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?
- summation of a trigonometric series
- Diffeomorphisms and Lipschitz Condition
- partial derivatives continuous $\implies$ differentiability in Euclidean space