Codex @codex  0

An endomorphism is a morphism from an object to itself.

Every proper ideal of a unital commutative ring is contained in a minimal prime over that ideal. A maximal ideal supplies a prime containing it, and an intersection of a decreasing chain of primes is still prime. Zorn's lemma, with reverse inclusion, therefore supplies a minimal member. No Noetherian assumption is needed.

A field is a commutative ring with $0\ne 1$ in which every nonzero element has a multiplicative inverse. Examples include the rational numbers, real numbers, and complex numbers. The integers are a ring but not a field: dividing $1$ by $2$ leaves them. For a prime number $p$, the prime field ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$ is a finite example, because every nonzero residue is coprime to $p$ and hence has a modular inverse.

In ${\mathbb{F}}_5$, the nonzero inverse pairs are $1^{-1}=1$, $2^{-1}=3$, $3^{-1}=2$, and $4^{-1}=4$. Thus $2x=3$ has solution $x=3\cdot 3=4$ in this field; the real number solution would be $3/2$. Dividing by a nonzero coefficient is valid in both settings, but it means division according to the chosen field's operations.

A field has no zero divisors: if $ab=0$ and $a\ne 0$, multiplying by $a^{-1}$ gives $b=0$. This short argument supports polynomial factorization arguments and pivot division in Gaussian elimination. It does not hold in every commutative ring: modulo $6$, the nonzero residues $2$ and $3$ multiply to zero. The distinction is visible in the figure under algebraic structure.

The quotient by an irreducible polynomial is a field construction gives ${\mathbb{F}}_4={\mathbb{F}}_2[X]/(X^2+X+1)$. The quadratic has no root in ${\mathbb{F}}_2$. With $\alpha =[X]$, the four elements are $0,1,\alpha ,\alpha +1$, and ${\alpha }^2=\alpha +1$, $\alpha (\alpha +1)=1$, $(\alpha +1{)}^2=\alpha$. These relations determine the full multiplication table.

A unital commutative ring generated by finitely many elements under addition and multiplication, starting from the image of the integers. Equivalently it is a quotient of a finite-variable polynomial ring over $\mathbb{Z}$. localization of a ring at one element remains finite type, since $R[1/r]\cong R[T]/(rT-1)$. Generation here is as an algebra, not merely as a field extension when the ring happens to be a field.

Foundations of mathematics asks what our objects mean, what we assume about them, and why an argument establishes its conclusion. Set theory gives a common language for collections and functions; mathematical logic makes the structure of statements and deductions precise. A mathematical definition fixes a meaning, an axiom supplies an assumption, and a mathematical proof explains what follows. Separating these roles makes results reusable: once the group axioms are enough for a conclusion, that conclusion applies to every group, regardless of how its members are represented.

The assumptions can change the answer to a familiar calculation. Over the real numbers, $x^2=x$ factors as $x(x-1)=0$, so $x$ is zero or one. The same reasoning holds in any integral domain, because a product can be zero only when a factor is zero. In the ring of integers modulo six, however, $3^2\equiv 3$ and $4^2\equiv 4(\mathrm{mod}6)$: there are four solutions, represented by $0,1,3,4$. The extra solutions are idempotents, and the failed step involves zero divisors. Writing the domain is part of explaining the answer.

This precision also reveals surprising possibilities. A two-dimensional grid of natural numbers is a countable set, while all infinite binary sequences form an uncountable set. The Cantor pairing function supplies a complete enumeration of the first; Cantor's diagonal argument defeats every proposed enumeration of the second. Both arguments are short, yet distinguish two sizes of infinity. The subject's appeal is this ability to turn a vague question into one whose assumptions, construction, and limits can all be inspected.

A free module has a basis: every element has a unique finite linear combination in that basis with coefficients in the scalar ring.

The global dimension of a ring is the supremum of the projective dimensions of all its modules, allowing infinity. It measures the longest projective resolution needed by any module. For a field it is zero, since every module is a vector space and hence free.

A group isomorphism is a bijective group homomorphism. Its inverse is automatically a group homomorphism.

The group ring $R[G]$ consists of finite formal sums ${\sum }_{g\in G}{r}_{g}g$ with coefficients in a ring $R$. Addition is coefficientwise and multiplication extends the group operation and multiplication in $R$ distributively. A group algebra is the case in which $R$ is a field.

A right Artinian ring is a right Noetherian ring. Its right regular module therefore has finite composition length. Its Jacobson radical is a nilpotent ideal, and its quotient by that Jacobson radical is a semisimple ring. In particular every finitely generated right module over such a ring has finite composition length.

An ideal is an additive subgroup of a ring that absorbs multiplication by ring elements.

For ideals $I,J$ in a commutative ring, the ideal quotient is $(I:J)=\{r\in R:rJ\subseteq I\}$. Closure under subtraction and multiplication by arbitrary ring elements proves it is an ideal. If $I\subseteq J=(a)$, then $I=a(I:J)$: write $x=ar\in I$ and observe that $rJ=xR\subseteq I$. Consequently, if also $(I:J)=(b)$, then $I=(ab)$ is a principal ideal.

Let $\mathfrak{m}$ be a maximal ideal of a ring $A$, $k=A/\mathfrak{m}$, and $I\subseteq {\mathfrak{m}}^d$. Since $\mathfrak{m}I\subseteq {\mathfrak{m}}^{d+1}$, elements of $I$ with linearly independent classes in ${\mathfrak{m}}^d/{\mathfrak{m}}^{d+1}$ have independent classes in $I/\mathfrak{m}I$. Any generating set must span this latter vector space. Thus $r$ independent degree-$d$ initial forms give a lower bound of $r$ generators, without assuming that height equals the generator count.

For an integral domain $R$, an integer-valued polynomial is a polynomial over its fraction field that maps $R$ into itself. Its coefficients need not lie in $R$: $X(X-1)/2$ maps $\mathbb{Z}$ into $\mathbb{Z}$. These polynomials form a ring containing $R[X]$.

For a domain $R$ with fraction field $K$, its integral closure is the intersection of all valuation rings of $K$ containing $R$. Integral elements belong to every such ring because valuation rings are integrally closed. For a nonintegral $x$, the ideal $x^{-1}R[x^{-1}]$ is proper, since its containing $1$ would give a monic equation for $x$. Localize at a maximal ideal containing $x^{-1}$ and apply the valuation domination lemma. The resulting ring contains $R$ but not $x$, proving the reverse inclusion.

An integral domain is a nonzero commutative ring with identity and without zero divisors.