Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 3 / b

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- Solution b

Solution

The set

I

contains zero. If

x^{m} \in I

, then

(- x)^{m} = (- 1)^{m} x^{m} \in I

, giving closure under additive inverses; also

(r x)^{m} = r^{m} x^{m} \in I

for every

r \in R

.

To prove closure under addition, let

x^{m}, y^{n} \in I

. Expand

(x + y)^{m + n - 1}

by the binomial theorem. In every monomial

x^{j} y^{m + n - 1 - j}

, either

j \geq m

or

m + n - 1 - j \geq n

. Thus each term contains a factor

x^{m}

or

y^{n}

and belongs to

I

. Hence

(x + y)^{m + n - 1} \in I

, so

x + y \in I

. The additive subgroup and absorption properties are proved, giving

I is an ideal of R .

This ideal is the radical of an ideal. The finite exponent for a sum uses commutativity of the ring.

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