Prove U=Z where addition is regular addition in Z, but multiplication is defined by a*b=a IS NOT A RING
$begingroup$
Ok,so clearly
a+b ∈ R
a+(b+c)=(a+b)+c
a+b = b+a
a+0 = a = 0+a, 0 ∈ R a ∈ R
a+x = 0, x = -a ∈R
now
a ∈ R b ∈ R and ab = a ∈R
a(bc) = a(b) = a
(ab)c = (a)c = a
a(b+c) = ab+ac = a+a
(a+b)c = ac+bc = a+b
Why is this not a ring? It seems to me that it satisfies all the axioms
abstract-algebra
asked Dec 9 '18 at 6:08
s_healys_healy
205
$endgroup$
add a comment |
$begingroup$
Ok,so clearly
a+b ∈ R
a+(b+c)=(a+b)+c
a+b = b+a
a+0 = a = 0+a, 0 ∈ R a ∈ R
a+x = 0, x = -a ∈R
now
a ∈ R b ∈ R and ab = a ∈R
a(bc) = a(b) = a
(ab)c = (a)c = a
a(b+c) = ab+ac = a+a
(a+b)c = ac+bc = a+b
Why is this not a ring? It seems to me that it satisfies all the axioms
abstract-algebra
asked Dec 9 '18 at 6:08
s_healys_healy
205
$endgroup$
4
$begingroup$
So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
$endgroup$
– Yadati Kiran
Dec 9 '18 at 6:13
$begingroup$
ohhhhhhhhhhhhh ok thank you
$endgroup$
– s_healy
Dec 9 '18 at 6:44
add a comment |
$begingroup$
Ok,so clearly
a+b ∈ R
a+(b+c)=(a+b)+c
a+b = b+a
a+0 = a = 0+a, 0 ∈ R a ∈ R
a+x = 0, x = -a ∈R
now
a ∈ R b ∈ R and ab = a ∈R
a(bc) = a(b) = a
(ab)c = (a)c = a
a(b+c) = ab+ac = a+a
(a+b)c = ac+bc = a+b
Why is this not a ring? It seems to me that it satisfies all the axioms
abstract-algebra
asked Dec 9 '18 at 6:08
s_healys_healy
205
$endgroup$
Ok,so clearly
a+b ∈ R
a+(b+c)=(a+b)+c
a+b = b+a
a+0 = a = 0+a, 0 ∈ R a ∈ R
a+x = 0, x = -a ∈R
now
a ∈ R b ∈ R and ab = a ∈R
a(bc) = a(b) = a
(ab)c = (a)c = a
a(b+c) = ab+ac = a+a
(a+b)c = ac+bc = a+b
Why is this not a ring? It seems to me that it satisfies all the axioms
abstract-algebra
abstract-algebra
asked Dec 9 '18 at 6:08
s_healys_healy
205
asked Dec 9 '18 at 6:08
s_healys_healy
205
asked Dec 9 '18 at 6:08
s_healys_healy
205
asked Dec 9 '18 at 6:08
s_healys_healy
205
asked Dec 9 '18 at 6:08
s_healys_healy
205
205
4
$begingroup$
So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
$endgroup$
– Yadati Kiran
Dec 9 '18 at 6:13
$begingroup$
ohhhhhhhhhhhhh ok thank you
$endgroup$
– s_healy
Dec 9 '18 at 6:44
add a comment |
4
$begingroup$
So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
$endgroup$
– Yadati Kiran
Dec 9 '18 at 6:13
$begingroup$
ohhhhhhhhhhhhh ok thank you
$endgroup$
– s_healy
Dec 9 '18 at 6:44
4
4
$begingroup$
So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
$endgroup$
– Yadati Kiran
Dec 9 '18 at 6:13
$begingroup$
So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
$endgroup$
– Yadati Kiran
Dec 9 '18 at 6:13
$begingroup$
ohhhhhhhhhhhhh ok thank you
$endgroup$
– s_healy
Dec 9 '18 at 6:44
$begingroup$
ohhhhhhhhhhhhh ok thank you
$endgroup$
– s_healy
Dec 9 '18 at 6:44
add a comment |
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4
$begingroup$
So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
$endgroup$
– Yadati Kiran
Dec 9 '18 at 6:13
$begingroup$
ohhhhhhhhhhhhh ok thank you
$endgroup$
– s_healy
Dec 9 '18 at 6:44