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2 votes

Find an irreducible polynomial in $\mathbb Q[x]$ of degree $726$.

The number 727 is prime. Hence, the cyclotomic polynomial $\Phi_{727}(X) = \frac{X^{727} - 1}{X - 1} \in \mathbb{Q}[X]$ of degree 726 is irreducible.
ByteBlitzer's user avatar
0 votes
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For what integers $m\gt n\gt 0$, the polynomial $x^m+x^n+1$ is irreducible over $\mathbb Q$?

I will reproduce here my answer [copied from Math Reviews] to https://mathoverflow.net/questions/56579/about-irreducible-trinomials (one should also see the other answer posted to that question, as ...
Gerry Myerson's user avatar
2 votes

Irreducibility over Field extensions.

This argument only works for $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}, ..., \sqrt{p_{k-1}})$. It uses concepts from algebraic number theory, which you may or may not have studied. Note that $p$ is ...
Lukas Heger's user avatar
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4 votes
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Is this $f(x) = x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$ irreducible in GF(5)?

If $f(x)$ is not irreducible, then it has an irreducible factor of degree $2$ or $3$ (it has no linear factors). Suppose $\alpha$ is a root of $f(x)$ in some extension $K$ of $\mathbb{F}_5$. Then $\...
Arturo Magidin's user avatar
5 votes

Is this $f(x) = x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$ irreducible in GF(5)?

This is true, because $f(x)=\frac{1-x^7}{1-x}$, that is the roots of $f$ are all $7$-th roots of $1$. Let $\alpha$ be a root of $f$, consider the Galois orbit $\alpha, \alpha^5, \alpha^{5^2}, \cdots$. ...
Just a user's user avatar
0 votes

Prove: either there exists a $ j $ such that $ |x_j| > 1 $, or $ |x_j| = 1 $ for all $ j $.

Note that from Newton's identities it follows that $\prod_{i=1}^n x_i$ is an integer to note that $\prod_{i=1}^n |x_i|\ge 1$. Now, note that either there exists a $j$ such that $|x_j|>1$ or $|x_j|\...
Samrat Mukhopadhyay's user avatar
0 votes

Determine all of the monic irreducible polynomials in $\mathbb Z_3 [x]$ of degree $4.$

Consider the tower $F_3\subset F_9\subset F_{81}$ of fields. There are $81-9=72$ elements $a$ that generate a degree 4 extension. Each $a$ gives an irreducible $f=f_{F_3}^a$ of degree 4 and each $f$ ...
Alex Heinis's user avatar
1 vote

For $k\subset F \subset E$ algebraic field extensions, if "all" irreducible polynomials with a root in E factor in $F[x]$ then $F=E$?

Another example: consider $k=\mathbb{Q}$, $F=\mathbb{Q}[\sqrt{2}]$, $E=\mathbb{Q}[\sqrt[4]{2}]$. The only field $L$ with $k\subseteq L\subseteq E$ and $[L:k]=2$ is $F$. This follows from Galois Theory:...
Arturo Magidin's user avatar
1 vote
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For $k\subset F \subset E$ algebraic field extensions, if "all" irreducible polynomials with a root in E factor in $F[x]$ then $F=E$?

$k=\mathbb{F}_p$, $F=\mathbb{F}_{p^2}$, $E=\mathbb{F}_{p^4}$ is a counterexample. Let $f \in k[x]$ be the minimal polynomial of $a \in F \setminus E$. Now $deg(f)$ divides 4, but if $deg(f) \leq 2$ ...
Alex Day's user avatar
  • 101
1 vote

What are examples of irreducible but not prime elements?

Here is a simple example: Let $k$ be a field, and consider the ring $R=k[x^2,xy,y^2]$. Consider the relation: $$(x^2)(y^2)=(xy)(xy)$$ $x^2,xy,y^2$ are irreducibles over $R$, however none of them are ...
cansomeonehelpmeout's user avatar
3 votes
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$x_1,...,x_4$ are irreducible in $k[x_1,..,x_4]/\langle x_1x_4-x_2x_3\rangle$

This looks like the coordinate ring of $2\times 2$ matrices with determinant $0$. Such matrices are of rank $1$, so can be written in a certain way, $x_1 = a b$, $x_4 = c d$, $x_2=a c$, $x_3 = b d$. ...
orangeskid's user avatar
  • 54.3k
0 votes

Irreducibility check for polynomials not satisfying Eisenstein Criterion.

The top answer gave a cubic example $x^3+x+1$. Here I try to find a quadratic example. In the end of this handout, Show that there are irreducible polynomials which Eisenstein cannot detect. More ...
hbghlyj's user avatar
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