71 votes

Irreducible polynomial which is reducible modulo every prime

If $-1$ is a square in $\Bbb F_p$ (which includes the case $p=2$), say $a^2=-1$, then we have $$X^4+1=X^4-a^2=(X^2+a)(X^2-a).$$ If $p$ is odd and $2$ is a square in $\Bbb F_p$, say $2=b^2$, then we ...
Hagen von Eitzen's user avatar
34 votes

No finite field is algebraically closed

Let's suppose that $ K $ is finite and write $ K=\{\alpha_{1}, \ldots , \alpha_{n}\}$. Now take the polynomial $ p (x)=(x-\alpha_{1})\ldots (x-\alpha_{n}) +1\in K[x]$. It's easy to see that $ p (x) $ ...
Xam's user avatar
  • 6,109
25 votes
Accepted

In a finite field product of non-square elements is a square

If the characteristic of $F$ is $2$, then $x\mapsto x^2$ is an automorphism of $F$ and therefore every element is a square. So we can assume the characteristic is an odd prime. Consider the ...
egreg's user avatar
  • 238k
25 votes
Accepted

Normed vector spaces over finite fields

There is a "standard" way to consider normed spaces over arbitrary fields but these are not well-behaved in the case of scalars in finite fields. If you want to work with norms on vector ...
Chilote's user avatar
  • 4,231
24 votes

are the integers modulo 4 a field?

No. Addition and multiplication mod $n$ are well defined, so $\mathbb{Z}/n$, the integers mod $n$, is always a ring, but not a field in general unless $n$ is a prime. In particular, the integers mod 4,...
ziggurism's user avatar
  • 16.7k
20 votes

Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$

For any distinct primes $p_1,p_2$ the polynomial $$x^4-2(p_1+p_2)x^2+(p_1-p_2)^2,(1)$$ is irreducible in $\Bbb Q$, but this polynomial is reducible modulo $p$ for any prime $p$. Let us see why: It is ...
Camilo Arosemena-Serrato's user avatar
19 votes

What is an extension field? Covered differently in math & in cryptography.

The grammar of "extension field" is that it takes as input two fields, a smaller field $F$ and a bigger field $K$ into which $F$ embeds, so that we can say "$K$ is an extension of $F$.&...
Qiaochu Yuan's user avatar
18 votes
Accepted

Cat quiz (to solve with the determinant)

Let $M$ be the $100 \times 100$ matrix over $\mathbb{F}_2$ that has, at index $(i, j)$, a $1$ if cat $i$ likes food brand $j$ and a $0$ otherwise. Now, I like to see the Leibniz expansion of the ...
Christopher Gadzinski's user avatar
17 votes
Accepted

Diagonalizability of symmetric bilinear forms over fields of characteristic $2$

... Hence, assume that $H\neq0$, then there exists $z \in V$ such that $H(z,z) \neq 0$. ... This is not true in characteristic $2$. Let $\Bbb F$ be any field of that characteristic, and, for example, ...
Travis Willse's user avatar
17 votes
Accepted

Squares which are not 1 + a square in finite fields of odd characteristic?

Short answer I found to the existence part of my question. Let $\lvert F \rvert = p^n$ for some odd prime $p$ and a natural number $n$. The number of squares in $F$ is given by $\frac{p^n + 1}{2}$, a ...
Bib-lost's user avatar
  • 3,880
16 votes

How to find minimal polynomial for an element in $\mbox{GF}(2^m)$?

This isn't too difficult because we only need methods from linear algebra. Let me do an example. I pick the field $GF(2^5)$ because for smaller fields I know the answer by heart, and I would fall back ...
Jyrki Lahtonen's user avatar
14 votes

Find the Galois group of $x^3-2$ over the field $\mathbb F_5$

We have that any irreducible polynomial $f(X)\in\mathbb{F}_q[X]$ of degree $n$ has the Galois group of $f$ over $\mathbb{F}_q$ cyclic of degree $n$, and the Galois group is generated by the Frobenius ...
Daniel Buck's user avatar
  • 3,554
13 votes

Why are the elements of a galois/finite field represented as polynomials?

The short answer is that you do not need to view the elements of finite fields as polynomials, but it simply is the most convenient presentation for many a purpose. The slightly longer answer is that ...
Jyrki Lahtonen's user avatar
13 votes
Accepted

Field with $125$ elements

If a cubic polynomial of $\mathbb{F}_5[x]$ is reducible, then it splits into a linear factor and a quadratic factor or into the product of three linear factors. Linear factors are very easy to test ...
davidlowryduda's user avatar
  • 91.2k
13 votes
Accepted

$p^{th}$ roots of a field with characteristic $p$

We are looking for a root of $x^p-\alpha$; the formal derivative of this polynomial is zero, which means that $x^p-\alpha$ has repeated roots. Indeed, if $K$ is an extension of $F$ where the ...
egreg's user avatar
  • 238k
13 votes
Accepted

How to show the only absolute value on a finite field is the trivial one.

Let $\mathbb K^*$ be the multiplicative group of non-zero elements in $\mathbb K$. Then, suppose that $|x| = L \neq 1,0$. Claim : Then, $x$ has infinite order in the group $\mathbb K^*$. This is ...
Sarvesh Ravichandran Iyer's user avatar
13 votes
Accepted

$x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$

Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field. The field with $p^k$ elements has a ...
Alon Amit's user avatar
  • 15.6k
13 votes
Accepted

If $x^3$ is a square, is $x$ a square?

$x=x^3/x^2$, so if $x^3=a^2$ then $x=(a/x)^2$.
Angina Seng's user avatar
12 votes

The number of subspaces over a finite field

You want to construct a subspace of dimension $k$, i.e. you want to find the number of ways you can choose $k$ independent vector out of a vector space of dimension $n$. First see that, no. of ...
User's user avatar
  • 2,928
12 votes

What is the algebraic closure of $\mathbb F_q$?

Given finite fields $\mathbb{F}_{p^m}$ and $\mathbb{F}_{p^n}$ with $\gcd(m, n) = 1$ then the compositum is the finite field $\mathbb{F}_{p^{mn}}$. This allows us to define the algebraic closure of $\...
Dietrich Burde's user avatar
12 votes

When is the Frobenius endomorphism an automorphism?

In a field $F$ with characteristic $p$, the map $x\mapsto x^p$ is a field endomorphism. Indeed, $(xy)^p=x^py^p$ and $$ (x+y)^p=\sum_{k=0}^p \binom{p}{k}x^ky^{p-k} =x^p+y^p $$ because, when $p$ is ...
egreg's user avatar
  • 238k
12 votes
Accepted

Does every polynomial over a finite field have a square root modulo an irreducible polynomial?

The quotient $K = \operatorname{GF}(2^m)[x]/(g)$ is again a finite field of characteristic $2$. The map $$ f \colon K \to K, \quad y \mapsto y^2 $$ is additive because $$ ...
Jendrik Stelzner's user avatar
12 votes
Accepted

Prove that $x^3+2y^3+4z^3\equiv6xyz \pmod{7} \Rightarrow x\equiv y\equiv z\equiv 0 \pmod{7}$

Someone played a dirty trick on you. The form $x^3-6xyz + 2y^3+4z^3$ is the “norm form” for the extension $\Bbb F_{7^3}\supset\Bbb F_7$. That is, if you take a generator $\zeta=\sqrt[3]2$ of the big ...
Lubin's user avatar
  • 62.7k
11 votes

Irreducible of $\mathbb Z/p\mathbb Z$.

Let $F_9$ be the field $9$ elements. If the polynomial is reducible, it will have a factor of degree $\leq 2$, hence a root in $F_9$. Therefore it is enough to show that it has no roots in $F_9$. ...
David's user avatar
  • 6,306
11 votes
Accepted

Every finite field is perfect

At the beginning of the proof, $p$ is defined as the characteristic of the field. The claim is that the Freshman's Dream holds for that power, not just any power.
RCT's user avatar
  • 2,797
11 votes

$p^{th}$ roots of a field with characteristic $p$

If $a^p = b^p$ then $a^p-b^p = (a-b)^p=0$, and since you are in a field this implies $a=b$. This shows that for a field of characteristic $p$ the map $a \to a^p$ is always injective, and an injective ...
Nate's user avatar
  • 11.2k
11 votes
Accepted

Proof of Cayley-Hamilton Theorem in infinite fields only?

You can apply the following powerful idea: think of Cayley-Hamilton as a statement about the "universal matrix," the one whose entries are indeterminates $x_{ij}$ living in a polynomial ring ...
Qiaochu Yuan's user avatar
11 votes

Does the field with 27 elements contain a subfield with 9 elements?

If $L/K$ is a field extension, then $L$ is a vector space over $K$. If $L$ and $K$ are finite, $L$ must be finite-dimensional over $K$. Let $K$ have $q$ elements and $L$ have dimension $n$ as a vector ...
Angina Seng's user avatar

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