# Abstract Nonsense

## Ideals and Homomorphisms

Point of Post: In this post we explore some of the relations between ideals and homomorphisms, things like preimages of ideals, and images of epimorphisms are ideals, etc. We will not discuss the isomorphism theorems in this post, that will be covered later.

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Motivation

So far in our discussion of ideals we’ve completely ignored the interaction between ideals and homomorphisms (except that every kernel is an ideal). What kind of relationships should we expecdt to have? Since ideals are supposed to play the role for rings that normal subgroups played for groups it seems natural to look at the facts that held there. Things such as the preimage of an ideal is an ideal, etc. We also explore some of the ramifcations of these relationships for morphisms between certain kinds of rings, in particular for morphisms $k\to k'$ for fields $k,k'$.

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Ideals and Homomorphisms

We begin with a generalization of the previously proven fact that for a ring map $R\to S$ one has that $\ker f$ is an ideal of $R$. In particular:

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Theorem: Let $R$ and $S$ be rings and $f:R\to S$ a ring homomorphism. If $\mathfrak{a}\in\mathfrak{L}(S)$ then $f^{-1}(\mathfrak{a})\in\mathfrak{L}(R)$. If $J$ is a left or right ideal then $f^{-1}(J)$ is a left or right ideal.

Proof: Let $J\in\mathfrak{L}_L(S)$. We know from basic group theory that $f^{-1}(J)\leqslant R$ and so it suffices to show that $f^{-1}(J)$ has the left absorption property. To see this let $y\in f^{-1}(J)$ and $x\in R$. Then, $f(xy)=f(x)f(y)\in J$ and so $xy\in f^{-1}(J)$. Since $x,y$ were arbitrary we have that $f^{-1}(J)$ is a left ideal. The exact same argument works to show that if $J\in\mathfrak{L}_R(S)$ then $f^{-1}(J)\in\mathfrak{L}_R(R)$. Combining these two results tells us that if $\mathfrak{a}\in\mathfrak{L}(S)$ then $f^{-1}(\mathfrak{a})\in\mathfrak{L}(R)$.

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Corollary: Let $D$ be a non-zero division ring and $R$ an arbitrary ring, then for any non-zero morphism $f:D\to R$ we have that $f$ is a monomorphism.

Proof: We merely note that $\ker f$ is a two-sided ideal of $D$ but, by previous theorem we know this means that either $\ker f=\{0\}$ or $\ker f=D$, since we assumed the latter wasn’t true we may conclude the former. But, by a previous characterization of monomorphisms we may conclude that $f$ is a monomorphism. $\blacksquare$

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A particular case of this which shall be useful to us later is the following:

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Theorem: Let $k$ and $k'$ be a fields and $f:k\to k'$ be a unital morphism, then $f$ is a monomorphism (an embedding).

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Perhaps an obvious theorem, but one we skipped when we discussed ideals and which is immediate from this theorem is:

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Theorem: Let $R$ be a ring and $R'$ a subring. Then, if $J$ is a left, right, or two-sided ideal of $R$ then $J\cap R'$ is a left, right, or two-sided ideal of $R'$.

Proof: We merely note that $J\cap R'$ is equal to the preimage of $J$ under the inclusion $R'\hookrightarrow R$. $\blacksquare$

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One of the key things to keep in mind when dealing with ideals (now this is a guiding principle, not a rule) is that they are the ‘normal subgroups’ for rings (kernels of ideals–we shall prove the unproven half of this shortly). Consequently, the theory of them is very similar as the above shows (the preimage of normal subgroups is a subgroup and the theorem concerning division rings applies more generally to the (yet to be defined) concept of a simple ring which corresponds to a simple group) and so the next theorem should not be a surprise since it has the same statement as a theorem concerning normal subgroups. Namely:

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Theorem: Let $R$ and $S$ be rings and $J$ a left, right, or two-sided ideal of $R$. Then, $f(J)$ is a left, right, or two-sided ideal of $\text{im } f$.

Proof: We know that $f(J)$ is a subring of $\text{im } f$ and so it suffices to show it has the absorption property. All three cases are done the same, so we assume that $J$ is a left ideal. Then, for any $f(x)\in f(J)$ and $f(y)\in\text{im }f$ we have $f(y)f(x)=f(yx)\in f(J)$. Since $f(x),f(y)$ were arbitrary the conclusion follows. $\blacksquare$

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So, what’s the problem with saying that $f(J)$ is an ideal of $S$ itself, not just $\text{im }f$? Let’s kill two birds with one stone, in particular let’s note that if $R$ is a ring with subring $R'$ and $\mathfrak{a}$ is an ideal of $R'$ it need not be an ideal of $R$. Showing a counterexample to this also provides a counterexample to why the image of an ideal need not be an ideal of the full codomain ring by considering the inclusion mapping $R'\hookrightarrow R$. So, what’s the obvious counterexample for an ideal of a subring need not be an ideal of the ambient ring? Well, the most obvious I can think of is that $2\mathbb{Z}$ is an ideal of $\mathbb{Z}$ but evidently not an ideal of $\mathbb{Q}$.

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

June 24, 2011 -

## 3 Comments »

1. […] that is an ideal of since it’s the image of under the canonical projection  and we know that ideals are preserved under epimorphisms. So, it makes sense to consider the quotient ring . […]

Pingback by Third Ring Isomorphism Theorem « Abstract Nonsense | July 1, 2011 | Reply

2. […] proof that the opposite inclusion is true we merely note that if then is an ideal of since we know the preimage of ideals under morphisms are ideals. Since we may thus conclude that and since […]

Pingback by Fourth Ring Isomorphism Theorem (Lattice Isomorphism Theorem) (Pt. I) « Abstract Nonsense | July 9, 2011 | Reply

3. […] of say are of the form is as follows. We first note that if is an ideal then is an ideal since we know ideals are preserved under epimorphisms. We thus have that . To prove the reverse inclusion we […]

Pingback by Product of Rings (Pt. I) « Abstract Nonsense | July 11, 2011 | Reply