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# 7.2: Homomorfismos de anillo

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Como vimos tanto con grupos como con acciones grupales, ¡vale la paga considerar la estructura preservando funciones!

##### Definición 7.1.0

Vamos$$R$$ and $$S$$ be rings. Then $$\phi: R\rightarrow S$$ is a homomorphism if:

1. $$\phi$$ is homomorphism of additive groups: $$\phi(a+b)=\phi(a)+\phi(b)$$, and
2. $$\phi$$ preserves multiplication: $$\phi(a\cdot b)=\phi(a)\cdot \phi(b)$$.

Si el homomorfismo es una biyección, entonces es un isomorfismo.

Ejemplos:
1. Tenemos el homomorfismo de inclusión$$\iota: \mathbb{Z}\rightarrow \mathbb{Q}$$, which just sets $$\iota(n)=n$$. This map clearly preserves both addition and multiplication.
2. Considera el mapa$$\phi: \mathbb{Z}\rightarrow \mathbb{Z}_n$$ sending $$k$$ to $$k%n$$. We've seen that this is a homomorphism of additive groups, and can easily check that multiplication is preserved. Indeed,
$$\phi(a)=\phi(1+1+\cdots+1)=\phi(1)+\phi(1)+\cdots+\phi(1)=a\phi(1)=a%n$$.

Observe que cada elemento en$$\mathbb{Z}$$ can be written as a sum of many copies of $$1$$. Then we were able to figure out what the homomorphism does simply by knowing $$\phi(1)$$. As an example, consider the map $$\rho:\mathbb{Z}\rightarrow \mathbb{Z}_5$$ sending $$k$$ to $$(2k)%n$$. (Thus, $$\rho(0)=0, \rho(3)=1$$.) This can be shown, using the same argument as above, to be a ring homomorphism.

3. El mapa de evaluación$$e_k$$ is a function from $$R[x]$$ to $$R$$. For any polynomial $$f\in R[x]$$ and $$k\in R$$, we set $$e_k(f)=f(k)$$. This is a ring homomorphism! Let $$f(x)=a_nx^n+\cdots a_0x^0$$, and $$g(x)=b_nx^n+\cdots b_0x^0$$, where the $$a_i, b_i\in R$$. (We'll also allow leading coefficients to be zero in order to make it easy to add $$f$$ and $$g$$ formally.) We then check the ring homomorphism conditions:

b. ya que sabemos que$$e_k$$ is an additive homomorphism, we only need to check that it is multiplicative on monomials. But that's easy:
$$\begin{eqnarray}e_k((ax^n) (bx^m)) &=& e_k(abx^{n+m})\\ &=& abk^{n+m}=e_k(ax^n)e_k(bx^m).\end{eqnarray}$$

Ejercicio 7.1.1
1. Demostrar que$$\rho:\mathbb{Z}\rightarrow \mathbb{Z}_5$$ defined by $$\rho(k)=(3k)%5$$ is a ring homomorphism. Find the kernel and image of $$\rho$$.
Demostrar que$$\rho:\mathbb{Z}\rightarrow \mathbb{Z}_6$$ defined by $$\rho(k)=(3k)%6$$ is a ring homomorphism. Find the kernel and image of $$\rho$$. Al igual que con los grupos, también tenemos productos directos de anillos.
##### Definición 7.1.2

Vamos$$R$$ and $$S$$ be rings. Define the direct product $$R\times S$$ as the set $$\{(r,s) \mid r\in R, s\in S \}$$ with coordinate-wise operations: $$(r_1, s_1)+(r_2, s_2)=(r_1+r_2, s_1+s_2)$$, and $$(r_1, s_1)\cdot (r_2, s_2)=(r_1\cdot r_2, s_1\cdot s_2)$$.

Por supuesto, se debe verificar que se trata de un anillo comprobando los axiomas del anillo.

##### Ejercicio 7.1.3
1. Demuéstralo para cualquier anillo$$R$$ and $$S$$ that the product $$R\times S$$ is a ring.
2. Mostrar que el mapa de inclusión$$\iota: R\rightarrow R\times S$$ given by $$\iota(r)=(r,0)$$ is a ring homomorphism.
3. Demostrar que la proyección$$\pi: R\times S\rightarrow R$$ given by $$\pi((r,s))=r$$ is a ring homomorphism.

## Una pequeña digresión sobre la relación entre la buena informática y las buenas matemáticas

Recordemos que cuando trabajamos con grupos el núcleo de un homomorfismo era bastante importante; el núcleo dio lugar a subgrupos normales, los cuales fueron importantes en la creación de grupos cocientes.

Para los homomorfismos de anillo, la situación es muy similar. El núcleo de un homomorfismo de anillo todavía se llama núcleo y da lugar a anillos de cociente. De hecho, básicamente vamos a recrear todos los teoremas y definiciones que usamos para grupos, pero ahora en el contexto de anillos. Conceptualmente, ya hemos hecho el trabajo duro.

En la programación informática, la gente suele hablar del principio DRY: No te repitas, es decir, que no debes escribir el mismo código más de una vez. La razón, en la informática, es que es más fácil corregir errores o hacer modificaciones si un determinado fragmento de código aparece en un lugar distinto.

En matemáticas, tenemos un principio similar: la generalización. Cuando te encuentras haciendo lo mismo en diferentes contextos, significa que está pasando algo más profundo, y que probablemente haya una prueba de cualquier teorema que estés reprobando que no importa tanto en el contexto. Sería bueno, por ejemplo, recordar un solo concepto para grupos de cocientes, anillos de cocientes, espacios vectoriales de cociente, y cualquier otra cosa, en lugar de una mezcolanza de casos específicos de la misma idea básica.

Para el juego de homomorfismos, granos y cocientes, la generalización implica teoría de categorías y propiedades universales. La teoría de categorías está un poco más allá del alcance de estas notas, pero es una parte esencial de las matemáticas modernas y sirve de puente entre muchos campos diferentes de estudio matemático.

## Subring, Kernel, Imagen, Cociente.

Comenzamos con algunas definiciones.

##### Definición 7.1.4

Vamos$$R$$ be a ring. A subset $$S$$ of $$R$$ is a subring if $$S$$ is itself a ring using the same operations as $$R$$. (We don't require that $$S$$ has a multiplicative identity, though.)

Por ejemplo, tome$$R[x]$$, the polynomial ring over $$R$$. The set of degree $$0$$ polynomials is closed under addition and multiplication; indeed, this set is just a copy of $$R$$. Thus, $$R$$ is a subring of $$R[x]$$.

Por otro lado, considere el conjunto de todos los polinomios de grado mayor o igual a 2 en$$\mathbb{Z}[x]$$, which we'll denote $$P_{\geq2}$$. This is closed under addition (the sum of two polynomials has degree equal to the max of their degrees), and is closed under multiplication (the degree of the product is the sum of the degrees). Thus, it is a subring. However, the multiplicative identity in $$R[x]$$ is $$1$$, which has degree 0. So there is no unit in $P_{\geq 2}. Otro ejemplo: Tomar$$2\mathbb{Z}\subset \mathbb{Z}$$, the set of even integers. This set is closed under addition and multiplication, and is thus a subring. (The sum and product of two even integers is still even.) However, the even integers don't have the number $$1$$, and so there is no unit in $$2\mathbb{Z}$$. ##### Vamos$$P^n_{\geq 2}$$ denote all polynomials in $$\mathbb{Z}_n[x]$$ with degree $$\geq 2$$. Is $$P^4_{\geq 2}$$ a subring of $$\mathbb{Z}_n[x]$$? Why or why not? ##### Definición 7.1.6 Vamos$$\phi: R\rightarrow S$$ be a ring homomorphism. The kernel of $$\phi$$ is $$\{r\in R\mid \phi(r)=0\}$$, which we also write as $$\phi^{-1}(0)$$. The image of $$\phi$$ is the set $$\{\phi(r) \mid r\in R\}$$, which we also write as $$\phi(R)$$. De inmediato tenemos lo siguiente. Proposición 7.1.17 Vamos$$\phi: R\rightarrow S$$ be a ring homomorphism. Then the kernel of $$\phi$$ is a subring of $$R$$ and the image of $$\phi$$ is a subring of $$S$$. Prueba 7.1.8 Desde$$\phi$$ is a homomorphism of commutative additive groups, we know that the kernel and image are closed under addition. The kernel is closed under multiplication, because if $$\phi(a)=\phi(b)=0$$, then $$\phi(ab)=\phi(a)\phi(b)=0$$. The image is closed because if $$x, y \in \phi(R)$$, then there exist $$a, b\in R$$ such that $$\phi(a)=x, \phi(b)=y$$. Then $$xy=\phi(a)\phi(b)=\phi(ab)\in \phi(R)$$. Así como los granos de homomorfismos grupales eran tipos especiales de subgrupos, los granos de homomorfismos de anillo son tipos especiales de sutraes. ##### Definición 7.1.9 Un subring$$I$$ of a ring $$R$$ is an ideal if for any $$x\in I, r\in R$$, $$rx\in I$$ and$xr\in I.

Proposición 7.1.10

Vamos$$K$$ be the kernel of a ring homomorphism $$\phi:R\rightarrow S$$. Then $$K$$ is an ideal.

Prueba 7.1.11

Para cualquier$$x\in K$$, we have $$\phi(x)=0$$. Then $$\phi(rx)=\phi(r)\phi(x)=\phi(r)0=0$$. Similarly, $$\phi(xr)=0$$. Thus, $$K$$ is a two-sided ideal.

Los ideales están jugando exactamente el mismo papel que los subgrupos normales en el contexto de grupos; de hecho, un ideal es un subgrupo normal del grupo aditivo del anillo. En particular, podemos formar cosets y considerar el cociente$$R/\mathord I$$. Since it's an additive group, cosets of an ideal $$I$$ are of the form $$r+I = \{r+x | x\in I \}$$.

Teorema 7.1.12

Si$$I$$ is an ideal, then $$R/\mathord I$$ is a ring.

Prueba 7.1.13

Sabemos que bajo adición$$R/\mathord I$$ is a commutative group. So we just need to show that the multiplication distributes over addition. For this we have:
$$( (r+I)+(q+I) )(s+I) = rs+qs+I = (r+I)(s+I) + (q+I)(s+I)$$.
One can also check that the multiplication is associative and commutative if $$R$$ is associative and commutative. Likewise, if $$R$$ has a unit, then $$1+I$$ acts as a unit in $$R/\mathord I$$.

Finalmente, tenemos el teorema del isomorfismo.

Teorema 7.1.14: Teorema del isomorfismo
Vamos$$R$$ and $$S$$ be rings, and $$\phi:R\rightarrow S$$ a homomorphism. Then the image of $$\phi$$ is isomorphic to $$R/\mathord I$$.
Prueba 7.1.15

Para probar el teorema del isomorfismo, construir un homomorfismo a partir de$$R/\mathord I$$ to the image of $$\phi$$, just as we did for groups, and show that it is a bijection.