A field is an integral domain
Dependencies:
If $R$ is a field then it is also an integral domain. This means $R$ has no zero-divisors, i.e. the product of two non-zero numbers is always non-zero.
Proof by contradiction
Assume that $R$ has zero-divisors. Let $a, b \in R-\{0\}$ be two zero-divisors of $R$.
Since $R$ is a field, $a$ and $b$ have multiplicative inverses. Let them be $a^{-1}$ and $b^{-1}$ respectively.
\[ 1 = (a^{-1}a)(bb^{-1}) = a^{-1}(ab)b^{-1} = a^{-1}0b^{-1} = 0 \]
This is a contradiction. Hence $R$ has no zero-divisors. Therefore, $R$ is an integral domain.
Dependency for:
- Every complex matrix has an eigenvalue
- All eigenvalues of a symmetric operator are real
- A matrix is full-rank iff its determinant is non-0
- Product of linear factors is a factor
- The ideal generated by an irreducible polynomial is maximal
- Polynomial division theorem
- F[x]/p(x): A ring
- The ring F[x]/p(x) is a field iff p is irreducible
- p(x)F[x] = F[x] iff p is a non-zero constant
Info:
- Depth: 3
- Number of transitive dependencies: 5