![]() Every analysis text I know which discusses fields includes $1 \neq 0$ among the field axioms. In the context of analysis courses, fields appear at the beginning but only in a very shallow way. Let me end with some comments about forgetting about the field axiom $1 \neq 0$. As Qiaochu Yuan likes to say, the zero ring is "too simple to be simple": including it as a field would screw all this up. In particular, since ideals are $R$-submodules of $R$, an ideal $I$ of $R$ is maximal iff $R/I$ is a field. It is easy to see that a commutative ring if simple iff it is a field. This means that $R$ is not the zero ring but the only ideals of $R$ are $(0)$ and $R$. Then there is the notion of a simple commutative ring, which is a ring such that $R$ is simple as an $R$-module. We also have the important notion of a simple module, which is a nonzero module with only the zero module as a submodule, and there is the important fact that if $N \subset M$ are $R$-modules, then $N$ is a maximal submodule of $M$ iff $M/N$ is a simple module. So unlike all other rings satisfying the remaining field axioms, there are no nontrivial vector spaces over the zero ring and hence no nontrivial linear algebra. (As an aside, the Morita theory in the commutative case simply says that two commutative rings with isomorphic module categories are isomorphic.) But the only module over the zero ring is the zero module. In particular the modern perspective on commutative rings is to focus attention on modules over the ring. The reasons for excluding the zero ring from being a field have already been well-described by others. So we have work to do before we know whether we can write down the quotient ring or not! In more complicated rings this becomes more difficult or practically impossible.) $I \times (I I) = I$ and $I \times I I \times I = I$, so $\forall x,y,z \in F, x(y z) = xy xz$īased on the above, $\/\langle a_1,\ldots,a_n \rangle$ if there is some $d \geq 2$ which divides each of $a_1,\ldots,a_n$. Distributivity of multiplication over addition.$\forall x, y, z \in F$, $(x \times y) \times z = I \times I = I$ and $x \times (y \times z) = I \times I = I$, so $(x \times y) \times z = x \times (y \times z)$ $\forall x, y \in F, x \times y = I = y \times x$ However, because the additive identity need not have a multiplicative inverse, this is a vacuous truth. ![]() $\forall x \in F, \exists y = I \in F: x \times y = I$. $\forall x \in F, \exists y = I \in F: x y = I$ $\forall x \in F, x \times I = x$, so $I$ is the multiplicative identity. Existence of mulitiplicative identity.$\forall x \in F$ (i.e., for $x=I$), $x I = x$, so $I$ is the additive identity. If $x, y \in F$, then $x = y = I$, so $x y = I I = I \in F$. Define addition and multiplication such that $I I=I$ and $I \times I=I$. Consider the set $F$ consisting of the single element $I$.
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