Section 3.3 Noether’s Isomorphism theorem
Guiding Questions.
In this section, we’ll seek to answer the questions:
If the only rings that existed were polynomial rings, familiar systems of numbers like \(\Z, \Q, \R, \C\text{,}\) and matrix rings, there would still be enough to justify defining the concept of a ring and exploring its properties. However, these are not the only rings that exist. In this section, we explore a way of building new rings from old by means of ideals. To better understand these new rings, we will also define two new classes of ideals: prime ideals, and maximal ideals. We end by briefly connecting these rings to a familiar problem from high school algebra.
As quotient rings provide fertile soil for building new examples of rings, it should not surprise us to find that homomorphisms interact with quotient rings in interesting and useful ways. Chief among them are the isomorphism theorems. In this subsection, we focus primarily on the First Isomorphism Theorem.
We have seen that any homomorphism \(\p : R\to S\) gives rise to an ideal of \(R\text{,}\) namely \(\ker\p\text{.}\) Our next theorem demonstrates that, given a commutative ring with identity \(R\text{,}\) every ideal is the kernel of some homomorphism defined on \(R\text{.}\)
Theorem 3.3.1.
Let \(R\) be commutative with identity and \(I\) an ideal of \(R\text{.}\) Define \(\p: R\to R/I\) by \(\p(r) = r+I\text{.}\) Then \(\p\) is a homomorphism with \(\ker\p = I\text{.}\)
In what follows, we work toward a proof of the First Isomorphism Theorem for Rings.
Throughout, let \(R\) and \(S\) be commutative rings with identity, and let \(\p : R\to S\) be a homomorphism. Recall that \(\im \p = \setof{s\in S}{\p(r) = s\text{ for some } r\in R}\text{.}\)
Define \(f: R/\ker \p \to \im \p\) by \(f(r+\ker \p) = \p(r)\text{.}\)
Lemma 3.3.2.
Using the notation from above,
\(f\) is a
well-defined function.
Lemma 3.3.3.
Using the notation above, \(f\) is a homomorphism.
Lemma 3.3.4.
Using the notation above, \(f\) is one-to-one.
Lemma 3.3.5.
Using the notation above, \(f\) is onto.
We thus obtain the following, due to Emmy Noether:
Theorem 3.3.6 (E. Noether). Noether’s Isomorphism Theorem.
Let \(\p : R\to S\) be a homomorphism of commutative rings. Then \(R/\ker \p \cong \im \p\text{.}\)
In particular, if \(\p : R\to S\) is onto, \(R/\ker \p \cong S\text{.}\)
The First Isomorphism Theorem gives a useful way of establishing an isomorphism between a quotient ring \(R/I\) and another ring \(S\text{:}\) find an onto homomorphism \(R\to S\) with kernel \(I\text{.}\)
Theorem 3.3.7.
We have the following isomorphisms of rings.
\(\displaystyle \Z/\ideal{m} \cong \Z_m\)
\(\displaystyle \Q[x]/\ideal{x-5} \cong \Q\)
\(\displaystyle \R[x]/\ideal{x^2+1} \cong \C\)
Activity 3.3.8.
Let \(R = \Z_6\) and define \(\p : \Z_6 \to \Z_2\) by \(\p(\overline{x}) = \overline{x}\text{.}\) That is, \(\p\) sends an equivalence class \(\overline{x}\in \Z_6\) represented by \(x\in \Z\) to the equivalence class represented by \(x\) in \(\Z_2\text{.}\)
Prove that \(\p\) is a homomorphism.
Is \(\p\) onto? Justify.
Compute \(\ker\p\) (that is, list the elements in the set). Is \(\p\) one-to-one?
Without appealing to the definition, is \(\ker\p\) prime? Maximal? Explain.
