# Tag Info

## Hot answers tagged finite-fields

### 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 ...
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### 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 $p (x)$ ...
• 6,109
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### 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 ...
• 238k
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### 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 ...
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### 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,...
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### 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 ...

### 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 $F$.&...
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### 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 ...
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### 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, ...
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### 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 ...
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### 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 ...
• 132k

### 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 ...
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### 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 ...
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### 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 ...
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### $p^{th}$ 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 ...
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### 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 ...
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### $x^4+x^3+x^2+x+1$ 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 ...
• 15.6k
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### If $x^3$ is a square, is $x$ a square?

$x=x^3/x^2$, so if $x^3=a^2$ then $x=(a/x)^2$.
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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 ...