14. IDEALS AND FACTOR RINGS
157
(6) Let Z[x] be the ring of all polynomials with integer coefficients and let I be
the subset of Z[x] of all polynomials with even constant term. Then
I = hx, 2i
is an ideal of Z[x].
Proof.
[To show I = hx, 2i.]
If f (x) 2 hx, 2i, f (x) = xg(x) + 2h(x) where g(x), h(x) 2 R. Then
f (0) = 0 · g(0) + 2 · h(0) = 2h(0),
so f (x) 2 I. Also, if f (x) 2 I,
f (x) = anxn + an 1xn
1
+ · · · a1x + 2k =
x(anxn
Thus, by mutual inclusion, I = hx, 2i.
1
+ an 1xn
2
+ · · · a1) + 2k 2 hx, 2i.
[Show I is an ideal.] Suppose k(x) 2 Z[x] and f (x) = xg(x) + 2h(x) 2 I.
Then
p(x) = f (x)k(x) = k(x)f (x) = xk(x)g(x) + 2k(x)h(x) 2 I.
Also, if f (x) = xf1(x) + 2f2(x) and g(x) = f (x) = xg1(x) + 2g2(x),
f (x)
g(x) = x (f1(x)
so I is an ideal by Theorem 14.1.
g1(x) + 2 (f2(x)
g2(x) 2 I,
⇤
(7) Let R be the ring of all real-valued functions of a real variable. Let S be
the subset of all di↵erentiable functions (this means for f 2 S, f 0(x) is defined
for all real x. S is not an ideal of R.
8
>
x>0
<1,
Let f (x) = 1 2 S and let g(x) = sgn x = 0,
x = 0.
>
: 1, x < 0
h(x) = g(x)f (x) = sgn x 62 S. Thus S is not an ideal, but is a subring of R.