Introduction to Ring Theory (4)

It’s now time to talk about homomorphisms.

Definition. Let R, S be rings. A function f : R → S is a ring homomorphism if it satisfies the following:

  • f(1R) = 1S;
  • f(x+y) = f(x) + f(y) for all x, y in R;
  • f(xy) = f(x) f(y) for all x, y in R.

Some immediate properties:

  • The second condition tells us f is a homomorphism of the underlying additive groups.
  • The first condition is not superfluous: consider the map R → R × R which takes x to (x, 0). This map satisfies the second and third conditions but not the first.
  • The first and third conditions tell us that units are taken to units, since if x is a unit, then xy = 1, so f(xf(y) = f(1) and f(x) is also a unit. This helps to explain why we need the first condition: so that units go to units under a homomorphism.

However, zero-divisors don’t go to zero-divisors in general. For example, consider the map R × R → R which takes (xy) to x. Then (1, 0) is a zero-divisor but the image, 1, is not.

One particularly important homomorphism is the projection R → R/I, for an ideal I of R, which takes x to x+I.

Needless to say, composition of two homomorphisms is also a homomorphism. One also defines a monomorphism (resp. epimorphismisomorphism) to be an injective (resp. surjective, bijective) homomorphism.

Effect on Subrings and Ideals

Let’s examine the effect of f and its inverse on subrings and ideals.

Theorem. Let f : R → S be a ring homomorphism. Then:

  • if R’ is a subring of R, then f(R’) is a subring of S;
  • if S’ is a subring of S, then f-1(S’) is a subring of R;
  • if J is an ideal of S, then f-1(J) is an ideal of R;
  • if I is an ideal of R, then f(I) is not an ideal of R in general.

The first three statements are easy to prove: once again, they illustrate the general philosophy that the “pullback” f-1 is better behaved that the “pushforward” f. We already saw this earlier with normal subgroups of groups. To obtain a counter-example for the fourth statement, take the inclusion map Z → Q. Then 2Z is an ideal in Z but clearly not an ideal in Q. [ Since Q is a field, its only ideals are 0 and itself. ]

In particular, we have the following special cases.

Definition. If f : R → S is a ring homomorphism, then f(R) =  im(f) is the image of f; this is a subring of S. On the other hand, f-1({0}) = ker(f) is the kernel of f; this is an ideal of R.

First Isomorphism Theorem

The three isomorphism theorems are strongly reminiscent of the case for groups.

First Theorem of Isomorphism. If f : R → S is a ring homomorphism and I = ker(f), then this induces an isomorphism φ:R/I → im(f)$ where φ(r+I) = f(r).

The proof is straight-forward: by the case for group theory, we already know that φ is an isomorphism of the additive groups. Clearly, it takes 1 to 1. To show it preserves product:

φ((r+I)(s+I)) = φ(rs+I) = f(rs) = f(r)f(s) = φ(r+I)φ(s+I).

One application of the theorem: suppose we wish to prove that if S is a subring of R, and I is an ideal of R, then ∩ I is an ideal of S. Indeed, the composed homomorphisms S → R → R/I has kernel \{s\in S: s+I = 0\} = S\cap I. Hence, the kernel of this map, ∩ I, is an ideal of S and we get an injection S/(∩ I) → R/I.

Another useful application: suppose we wish to prove that I is an ideal of R and R/I\cong S, where IR and S are explicitly defined. An easy way to do this is via constructing an epimorphism R → S with kernel I and letting the first isomorphism theorem do its job. Here’s a concrete example.

Example

Let’s take an example in the previous post, where U is the ring of 2 × 2 upper-triangular real matrices \begin{pmatrix}* & *\\{0}& *\end{pmatrix}. Map f : U → R×R by taking the two diagonal entries ab to the ordered pair (ab). Then f is an epimorphism so  we get

\mathbf{R}\times\mathbf{R} \cong U/\text{ker}(f),

where I = ker(f) is the set of all strictly upper-triangular real matrices \begin{pmatrix}{0}&*\\{0}&{0}\end{pmatrix}.

Second + Third Isomorphism Theorems

In preparing for the second isomorphism theorem, let’s prove that if S is a subring of R and I is an ideal of R, then S+I is a subring of R. Indeed, S+I is an additive subgroup of R, so it suffices to show that it’s closed under multiplication, but this follows from:

x_1, x_2\in S, y_1, y_2\in I \implies (x_1 + y_1)(x_2 + y_2) = \overbrace{x_1 x_2}^{\in S} + \overbrace{x_1 y_2 + y_1 x_2 + y_1 y_2}^{\in I}.

Another way of proving this would be:

(S+I)(S+I) = \overbrace{S^2}^{\subseteq S} + \overbrace{SI + IS + I^2}^{\subseteq I}\subseteq S + I.

The second method is actually “preferred” since it’s more concise: the reader is urged to get comfortable with that kind of notation. Just a reminder: the product AB of two sets here refers to the set of all finite sums \sum_{i=1}^n x_i y_i, with x_i\in A, y_i\in B.

Second Isomorphism Theorem. Let S\subseteq R and I\subseteq R be a subring and an ideal respectively. We already know that S+I \subseteq R is a subring and I\cap S\subseteq S is an ideal of S. Then:

S/(S\cap I) \cong (S+I)/I.

Once again, by the second isomorphism theorem for groups, we already have an isomorphism of additive groups. So it remains to consider whether the map preserves the unity “1” and the product of two elements. But the map takes s+(∩ I) on the LHS to s+I on the RHS. Thus it’s easy to verify that the map takes “1” to “1”, and preserves the product.

Finally, to prepare for the last isomorphism theorem, consider I\subseteq J \subseteq R, where I and J are ideals of R. Then the additive group quotient J/I is an ideal of R/I, because every element of J/I can be written as x+I for some x in J. Thus the products (x+I)(r+I) = xr+I and (r+I)(x+I) = rx+I are both elements of J/I since xr and rx are in J.

Third Theorem of Isomorphism. Let I\subseteq J \subseteq R where I and J are ideals of R. Then we have an isomorphism of rings:

(R/I) / (J/I) \cong R/J.

Same proof: by the results of group theory, we already know the underlying additve groups are isomorphism. To show isomorphism as rings, it remains to show 1 maps to 1, and product on the LHS maps to product on the RHS, which is quite easy.

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