Problem:
Solution:
For part (a), we show the set is closed from addition and multiplication.
Suppose $ f, g $ is in $ I(S) $, $ h \in k[x_1, \cdots , x_n] $ then for all $ x \in S $, $ f(x) = g(x) = 0 $.
Now $ (f + g)(x) = 0 $ and also $ (hg)(x) = 0 $, so $ f + g $ and $ hg $ are both in $ I(S) $.
Therefore $ I(S) $ is an ideal.
For part (b), we do exactly the same as in Exercise 10. Now we have $ f(x, y) = h(x, y)(x - y) + r(x) $, $ f(a, a) = 0 \implies r(a) = 0 $, $ \forall a \neq 1 $. so now we still have $ r(x) = 0 $ and therefore the ideal is still $ \langle x - y \rangle $.
For part (c), we already know polynomials that vanish on the whole integer grid is necessarily the zero polynomial, so $ I(\mathbf{Z}) $ is simply $ \{ 0 \} $.
Solution:
For part (a), we show the set is closed from addition and multiplication.
Suppose $ f, g $ is in $ I(S) $, $ h \in k[x_1, \cdots , x_n] $ then for all $ x \in S $, $ f(x) = g(x) = 0 $.
Now $ (f + g)(x) = 0 $ and also $ (hg)(x) = 0 $, so $ f + g $ and $ hg $ are both in $ I(S) $.
Therefore $ I(S) $ is an ideal.
For part (b), we do exactly the same as in Exercise 10. Now we have $ f(x, y) = h(x, y)(x - y) + r(x) $, $ f(a, a) = 0 \implies r(a) = 0 $, $ \forall a \neq 1 $. so now we still have $ r(x) = 0 $ and therefore the ideal is still $ \langle x - y \rangle $.
For part (c), we already know polynomials that vanish on the whole integer grid is necessarily the zero polynomial, so $ I(\mathbf{Z}) $ is simply $ \{ 0 \} $.
No comments:
Post a Comment