Heuristic in quotient rings with linear relations?
$begingroup$
Artin Algebra Chapter 11. Some solutions are here and here.
It looks like all I have to do is find an integer that equals $0$ if the answer turns out to be $mathbb Z/(n)$, which is the case for the exercises above. I get
(a) Multiply 3 to get 15=18. $n=3$ (b) Multiply 2 to get 20=6. $n=14$ (3) Factor $alpha^2+alpha$ to get 1=0. $n=1$
What is the basis for this heuristic? Is this merely a way to guess? Is this a coincidence?
How can I do it for the exercises below (c) and (d)? (Adjoining elements is not introduced until the next section)
(c) $6=0$, but $2alpha-1=0$.
It turns out we're not in $mathbb Z/(6)$ but in $mathbb Z/(n)$ for some $n$ where $6=0$, $n=1,2,3,6$.
Multiplying 3 I get $6alpha-3=0=-3$. Then $n=3$.
It turns out $n=14$ in 5.4(b) was lucky and so actually requires more proof?
(d) How do I do similarly to show $7=0$ for $2alpha^2-4=0=4alpha-5$?
abstract-algebra number-theory field-theory modular-arithmetic ideals
$endgroup$
add a comment |
$begingroup$
Artin Algebra Chapter 11. Some solutions are here and here.
It looks like all I have to do is find an integer that equals $0$ if the answer turns out to be $mathbb Z/(n)$, which is the case for the exercises above. I get
(a) Multiply 3 to get 15=18. $n=3$ (b) Multiply 2 to get 20=6. $n=14$ (3) Factor $alpha^2+alpha$ to get 1=0. $n=1$
What is the basis for this heuristic? Is this merely a way to guess? Is this a coincidence?
How can I do it for the exercises below (c) and (d)? (Adjoining elements is not introduced until the next section)
(c) $6=0$, but $2alpha-1=0$.
It turns out we're not in $mathbb Z/(6)$ but in $mathbb Z/(n)$ for some $n$ where $6=0$, $n=1,2,3,6$.
Multiplying 3 I get $6alpha-3=0=-3$. Then $n=3$.
It turns out $n=14$ in 5.4(b) was lucky and so actually requires more proof?
(d) How do I do similarly to show $7=0$ for $2alpha^2-4=0=4alpha-5$?
abstract-algebra number-theory field-theory modular-arithmetic ideals
$endgroup$
add a comment |
$begingroup$
Artin Algebra Chapter 11. Some solutions are here and here.
It looks like all I have to do is find an integer that equals $0$ if the answer turns out to be $mathbb Z/(n)$, which is the case for the exercises above. I get
(a) Multiply 3 to get 15=18. $n=3$ (b) Multiply 2 to get 20=6. $n=14$ (3) Factor $alpha^2+alpha$ to get 1=0. $n=1$
What is the basis for this heuristic? Is this merely a way to guess? Is this a coincidence?
How can I do it for the exercises below (c) and (d)? (Adjoining elements is not introduced until the next section)
(c) $6=0$, but $2alpha-1=0$.
It turns out we're not in $mathbb Z/(6)$ but in $mathbb Z/(n)$ for some $n$ where $6=0$, $n=1,2,3,6$.
Multiplying 3 I get $6alpha-3=0=-3$. Then $n=3$.
It turns out $n=14$ in 5.4(b) was lucky and so actually requires more proof?
(d) How do I do similarly to show $7=0$ for $2alpha^2-4=0=4alpha-5$?
abstract-algebra number-theory field-theory modular-arithmetic ideals
$endgroup$
Artin Algebra Chapter 11. Some solutions are here and here.
It looks like all I have to do is find an integer that equals $0$ if the answer turns out to be $mathbb Z/(n)$, which is the case for the exercises above. I get
(a) Multiply 3 to get 15=18. $n=3$ (b) Multiply 2 to get 20=6. $n=14$ (3) Factor $alpha^2+alpha$ to get 1=0. $n=1$
What is the basis for this heuristic? Is this merely a way to guess? Is this a coincidence?
How can I do it for the exercises below (c) and (d)? (Adjoining elements is not introduced until the next section)
(c) $6=0$, but $2alpha-1=0$.
It turns out we're not in $mathbb Z/(6)$ but in $mathbb Z/(n)$ for some $n$ where $6=0$, $n=1,2,3,6$.
Multiplying 3 I get $6alpha-3=0=-3$. Then $n=3$.
It turns out $n=14$ in 5.4(b) was lucky and so actually requires more proof?
(d) How do I do similarly to show $7=0$ for $2alpha^2-4=0=4alpha-5$?
abstract-algebra number-theory field-theory modular-arithmetic ideals
abstract-algebra number-theory field-theory modular-arithmetic ideals
asked Dec 20 '18 at 6:37
user198044
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$begingroup$
For 5.4.(a), (b), (c); it can be a way to guess the solution, but then you still have to prove that it works.
The fact that it works for these is not a coincidence, but the fact that each time, modulo the $n$ that you find, $alpha$ is some integer, so you "haven't actually adjoined anything".
Indeed for (a), $3=0$, and $2alpha =6$ so $2alpha =0$ and so $alpha = 3alpha -2alpha = 0$.
For (b), you have $alpha =10$ that's given; and for (c) there's only one ring in which $1=0$.
For 4.3, you need to be careful : you won't always be in the situation where $alpha$ (the class of $x$) is actually an integer; so you may have to deal with extensions of $mathbb{Z}/(n)$
answered Dec 20 '18 at 9:16
MaxMax
15k11143
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$begingroup$
For 5.4.(a), (b), (c); it can be a way to guess the solution, but then you still have to prove that it works.
The fact that it works for these is not a coincidence, but the fact that each time, modulo the $n$ that you find, $alpha$ is some integer, so you "haven't actually adjoined anything".
Indeed for (a), $3=0$, and $2alpha =6$ so $2alpha =0$ and so $alpha = 3alpha -2alpha = 0$.
For (b), you have $alpha =10$ that's given; and for (c) there's only one ring in which $1=0$.
For 4.3, you need to be careful : you won't always be in the situation where $alpha$ (the class of $x$) is actually an integer; so you may have to deal with extensions of $mathbb{Z}/(n)$
answered Dec 20 '18 at 9:16
MaxMax
15k11143
$endgroup$
add a comment |
$begingroup$
For 5.4.(a), (b), (c); it can be a way to guess the solution, but then you still have to prove that it works.
The fact that it works for these is not a coincidence, but the fact that each time, modulo the $n$ that you find, $alpha$ is some integer, so you "haven't actually adjoined anything".
Indeed for (a), $3=0$, and $2alpha =6$ so $2alpha =0$ and so $alpha = 3alpha -2alpha = 0$.
For (b), you have $alpha =10$ that's given; and for (c) there's only one ring in which $1=0$.
For 4.3, you need to be careful : you won't always be in the situation where $alpha$ (the class of $x$) is actually an integer; so you may have to deal with extensions of $mathbb{Z}/(n)$
answered Dec 20 '18 at 9:16
MaxMax
15k11143
$endgroup$
add a comment |
$begingroup$
For 5.4.(a), (b), (c); it can be a way to guess the solution, but then you still have to prove that it works.
The fact that it works for these is not a coincidence, but the fact that each time, modulo the $n$ that you find, $alpha$ is some integer, so you "haven't actually adjoined anything".
Indeed for (a), $3=0$, and $2alpha =6$ so $2alpha =0$ and so $alpha = 3alpha -2alpha = 0$.
For (b), you have $alpha =10$ that's given; and for (c) there's only one ring in which $1=0$.
For 4.3, you need to be careful : you won't always be in the situation where $alpha$ (the class of $x$) is actually an integer; so you may have to deal with extensions of $mathbb{Z}/(n)$
answered Dec 20 '18 at 9:16
MaxMax
15k11143
$endgroup$
For 5.4.(a), (b), (c); it can be a way to guess the solution, but then you still have to prove that it works.
The fact that it works for these is not a coincidence, but the fact that each time, modulo the $n$ that you find, $alpha$ is some integer, so you "haven't actually adjoined anything".
Indeed for (a), $3=0$, and $2alpha =6$ so $2alpha =0$ and so $alpha = 3alpha -2alpha = 0$.
For (b), you have $alpha =10$ that's given; and for (c) there's only one ring in which $1=0$.
For 4.3, you need to be careful : you won't always be in the situation where $alpha$ (the class of $x$) is actually an integer; so you may have to deal with extensions of $mathbb{Z}/(n)$
answered Dec 20 '18 at 9:16
MaxMax
15k11143
answered Dec 20 '18 at 9:16
MaxMax
15k11143
answered Dec 20 '18 at 9:16
MaxMax
15k11143
answered Dec 20 '18 at 9:16
MaxMax
15k11143
15k11143
add a comment |
add a comment |
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