Mixing
Describe $mathbb{Q}[x] / (x^2 + x)$ and $mathbb{Z}[x]/ (x-2, x^2+1)$ in simpler terms.
$begingroup$
I am trying to review for a test. One of the questions is asking me to find a simpler description of these two rings and I am not sure what they mean by this. Any suggestions?
abstract-algebra
asked Jan 2 at 1:30
user8513188user8513188
1096
$endgroup$
add a comment |
$begingroup$
I am trying to review for a test. One of the questions is asking me to find a simpler description of these two rings and I am not sure what they mean by this. Any suggestions?
abstract-algebra
asked Jan 2 at 1:30
user8513188user8513188
1096
$endgroup$
$begingroup$
For example, $(x-2,x^2+1) superseteq (x-2,x^2+1-(x-2)*x)=(x-2,1+4x)$. This is because $(x-2)$ being in the ideal means that $(x-2)*x$ is as well and we can take differences too. If you can use the ideal rules to go backwards as well you can get the other containment. Keep doing this until you get something simpler.
$endgroup$
– AHusain
Jan 2 at 1:53
$begingroup$
For the first one, just think about the definition of quotient ring in general. If $f,ginmathbb{Q}[x]/(x^2+x),$ they are the same (belong to the same equivalence class) iff $x^2+x|f-g.$ In particular, this means that there are only degree $0$ and $1$ polynomials in them. Now, you should also prove that any such polynomial is also in there too
$endgroup$
– dezdichado
Jan 2 at 1:57
$begingroup$
For the first one, factor and use CRT. For the second, note that $x=2$.
$endgroup$
– jgon
Jan 2 at 3:03
$begingroup$
Hey, @AHusain, the LaTeX you want is "supseteq", not "superseteq". (I had to look it up myself.)
$endgroup$
– JonathanZ
Jan 2 at 3:55
add a comment |
$begingroup$
I am trying to review for a test. One of the questions is asking me to find a simpler description of these two rings and I am not sure what they mean by this. Any suggestions?
abstract-algebra
asked Jan 2 at 1:30
user8513188user8513188
1096
$endgroup$
I am trying to review for a test. One of the questions is asking me to find a simpler description of these two rings and I am not sure what they mean by this. Any suggestions?
abstract-algebra
abstract-algebra
asked Jan 2 at 1:30
user8513188user8513188
1096
asked Jan 2 at 1:30
user8513188user8513188
1096
asked Jan 2 at 1:30
user8513188user8513188
1096
asked Jan 2 at 1:30
user8513188user8513188
1096
asked Jan 2 at 1:30
user8513188user8513188
1096
1096
$begingroup$
For example, $(x-2,x^2+1) superseteq (x-2,x^2+1-(x-2)*x)=(x-2,1+4x)$. This is because $(x-2)$ being in the ideal means that $(x-2)*x$ is as well and we can take differences too. If you can use the ideal rules to go backwards as well you can get the other containment. Keep doing this until you get something simpler.
$endgroup$
– AHusain
Jan 2 at 1:53
$begingroup$
For the first one, just think about the definition of quotient ring in general. If $f,ginmathbb{Q}[x]/(x^2+x),$ they are the same (belong to the same equivalence class) iff $x^2+x|f-g.$ In particular, this means that there are only degree $0$ and $1$ polynomials in them. Now, you should also prove that any such polynomial is also in there too
$endgroup$
– dezdichado
Jan 2 at 1:57
$begingroup$
For the first one, factor and use CRT. For the second, note that $x=2$.
$endgroup$
– jgon
Jan 2 at 3:03
$begingroup$
Hey, @AHusain, the LaTeX you want is "supseteq", not "superseteq". (I had to look it up myself.)
$endgroup$
– JonathanZ
Jan 2 at 3:55
add a comment |
$begingroup$
For example, $(x-2,x^2+1) superseteq (x-2,x^2+1-(x-2)*x)=(x-2,1+4x)$. This is because $(x-2)$ being in the ideal means that $(x-2)*x$ is as well and we can take differences too. If you can use the ideal rules to go backwards as well you can get the other containment. Keep doing this until you get something simpler.
$endgroup$
– AHusain
Jan 2 at 1:53
$begingroup$
For the first one, just think about the definition of quotient ring in general. If $f,ginmathbb{Q}[x]/(x^2+x),$ they are the same (belong to the same equivalence class) iff $x^2+x|f-g.$ In particular, this means that there are only degree $0$ and $1$ polynomials in them. Now, you should also prove that any such polynomial is also in there too
$endgroup$
– dezdichado
Jan 2 at 1:57
$begingroup$
For the first one, factor and use CRT. For the second, note that $x=2$.
$endgroup$
– jgon
Jan 2 at 3:03
$begingroup$
Hey, @AHusain, the LaTeX you want is "supseteq", not "superseteq". (I had to look it up myself.)
$endgroup$
– JonathanZ
Jan 2 at 3:55
$begingroup$
For example, $(x-2,x^2+1) superseteq (x-2,x^2+1-(x-2)*x)=(x-2,1+4x)$. This is because $(x-2)$ being in the ideal means that $(x-2)*x$ is as well and we can take differences too. If you can use the ideal rules to go backwards as well you can get the other containment. Keep doing this until you get something simpler.
$endgroup$
– AHusain
Jan 2 at 1:53
$begingroup$
For example, $(x-2,x^2+1) superseteq (x-2,x^2+1-(x-2)*x)=(x-2,1+4x)$. This is because $(x-2)$ being in the ideal means that $(x-2)*x$ is as well and we can take differences too. If you can use the ideal rules to go backwards as well you can get the other containment. Keep doing this until you get something simpler.
$endgroup$
– AHusain
Jan 2 at 1:53
$begingroup$
For the first one, just think about the definition of quotient ring in general. If $f,ginmathbb{Q}[x]/(x^2+x),$ they are the same (belong to the same equivalence class) iff $x^2+x|f-g.$ In particular, this means that there are only degree $0$ and $1$ polynomials in them. Now, you should also prove that any such polynomial is also in there too
$endgroup$
– dezdichado
Jan 2 at 1:57
$begingroup$
For the first one, just think about the definition of quotient ring in general. If $f,ginmathbb{Q}[x]/(x^2+x),$ they are the same (belong to the same equivalence class) iff $x^2+x|f-g.$ In particular, this means that there are only degree $0$ and $1$ polynomials in them. Now, you should also prove that any such polynomial is also in there too
$endgroup$
– dezdichado
Jan 2 at 1:57
$begingroup$
For the first one, factor and use CRT. For the second, note that $x=2$.
$endgroup$
– jgon
Jan 2 at 3:03
$begingroup$
For the first one, factor and use CRT. For the second, note that $x=2$.
$endgroup$
– jgon
Jan 2 at 3:03
$begingroup$
Hey, @AHusain, the LaTeX you want is "supseteq", not "superseteq". (I had to look it up myself.)
$endgroup$
– JonathanZ
Jan 2 at 3:55
$begingroup$
Hey, @AHusain, the LaTeX you want is "supseteq", not "superseteq". (I had to look it up myself.)
$endgroup$
– JonathanZ
Jan 2 at 3:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I find the easiest setting to think about questions like these are by using the first isomorphism theorem: if $phi$ is a ring homomorphism from $R$ onto $S$, then $S simeq R/ker(phi)$. This means you have to have some suspicion about what the result will look like, define a homomorphism into that ring, show that it is onto, and then prove that the kernel is what you expected. (This last step is usually done in two parts, establishing $I = ker(phi)$ via $I subseteq ker(phi)$ and $I supseteqker(phi)$, where $I$ is your desired ideal.)
The other useful hint here is that some of the easiest isomorphisms defined on $R[x]$ are the "point evaluations", i.e. substitute a fixed value for $x$ in each polynomial.
To illustrate this let's work out a very simple example showing that $R[x]/langle x rangle simeq R$ for any ring $R$. Define $phi : R[x] rightarrow R$ by $ phi: p(x) mapsto p(0)$. It's a homomorphism because of the way the ring operations are defined in $R[x]$. It is onto because for any $a in R$ we can define $p(x) = x + a in R[x]$ and $phi(p) = p(0) = a$. If $p(x) = x$ then $phi(p) = 0$ so $langle x rangle subseteq ker(phi)$. To prove the other inclusion, assume $q(x) in ker(phi)$, so $q(0) = 0$, i.e. the constant term of $q$ is zero, so we can factor an $x$ out, i.e. we can write $q$ as $xcdot p(x)$ for some $p$, so $q in langle x rangle$, and this shows $ker(phi) subseteq langle x rangle$. QED.
Notice that when we say we want to take $R[x]$ and mod out by $langle x rangle$ we can think of this as wanting to "set x equal to zero", To test your understanding, if we were given $R[x]/langle x-1 rangle$ we would want to set $x-1$ equal to zero. Which point evaluation would let us do that?
Now in your first problem you want to set $x^2 + x$ equal to zero. It's not quite as simple as the $R[x]/langle xrangle simeq R$ example, but you can think of this as setting $x(x+1)$ equal to zero, so try to combine "setting $x$ equal to zero" and "setting $x+1$ equal to zero" to get a complete answer.
For your second problem you should be able to pick the appropriate point evaluation to handle "setting $x-2$ to zero" as your first step. But then you still have to handle the $x^2 + 1$, or more accurately, whatever $x^2+1$ maps to after you've applied your first homomorphism. Go ahead and figure out what your point evaluation, $phi$, should be, and then tell yourself "Well, I want $phi(x^2+1)$ to be zero", and see if that suggests anything to you.
(I've written this as a bunch of hints to help you get started and guide your thinking, so there are still a bunch of details for you to figure out and nail down. Feel free to follow up with what you figure out and requests if you feel you need further hints.)
answered Jan 2 at 3:36
JonathanZJonathanZ
2,119613
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add a comment |
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$begingroup$
I find the easiest setting to think about questions like these are by using the first isomorphism theorem: if $phi$ is a ring homomorphism from $R$ onto $S$, then $S simeq R/ker(phi)$. This means you have to have some suspicion about what the result will look like, define a homomorphism into that ring, show that it is onto, and then prove that the kernel is what you expected. (This last step is usually done in two parts, establishing $I = ker(phi)$ via $I subseteq ker(phi)$ and $I supseteqker(phi)$, where $I$ is your desired ideal.)
The other useful hint here is that some of the easiest isomorphisms defined on $R[x]$ are the "point evaluations", i.e. substitute a fixed value for $x$ in each polynomial.
To illustrate this let's work out a very simple example showing that $R[x]/langle x rangle simeq R$ for any ring $R$. Define $phi : R[x] rightarrow R$ by $ phi: p(x) mapsto p(0)$. It's a homomorphism because of the way the ring operations are defined in $R[x]$. It is onto because for any $a in R$ we can define $p(x) = x + a in R[x]$ and $phi(p) = p(0) = a$. If $p(x) = x$ then $phi(p) = 0$ so $langle x rangle subseteq ker(phi)$. To prove the other inclusion, assume $q(x) in ker(phi)$, so $q(0) = 0$, i.e. the constant term of $q$ is zero, so we can factor an $x$ out, i.e. we can write $q$ as $xcdot p(x)$ for some $p$, so $q in langle x rangle$, and this shows $ker(phi) subseteq langle x rangle$. QED.
Notice that when we say we want to take $R[x]$ and mod out by $langle x rangle$ we can think of this as wanting to "set x equal to zero", To test your understanding, if we were given $R[x]/langle x-1 rangle$ we would want to set $x-1$ equal to zero. Which point evaluation would let us do that?
Now in your first problem you want to set $x^2 + x$ equal to zero. It's not quite as simple as the $R[x]/langle xrangle simeq R$ example, but you can think of this as setting $x(x+1)$ equal to zero, so try to combine "setting $x$ equal to zero" and "setting $x+1$ equal to zero" to get a complete answer.
For your second problem you should be able to pick the appropriate point evaluation to handle "setting $x-2$ to zero" as your first step. But then you still have to handle the $x^2 + 1$, or more accurately, whatever $x^2+1$ maps to after you've applied your first homomorphism. Go ahead and figure out what your point evaluation, $phi$, should be, and then tell yourself "Well, I want $phi(x^2+1)$ to be zero", and see if that suggests anything to you.
(I've written this as a bunch of hints to help you get started and guide your thinking, so there are still a bunch of details for you to figure out and nail down. Feel free to follow up with what you figure out and requests if you feel you need further hints.)
answered Jan 2 at 3:36
JonathanZJonathanZ
2,119613
$endgroup$
add a comment |
$begingroup$
I find the easiest setting to think about questions like these are by using the first isomorphism theorem: if $phi$ is a ring homomorphism from $R$ onto $S$, then $S simeq R/ker(phi)$. This means you have to have some suspicion about what the result will look like, define a homomorphism into that ring, show that it is onto, and then prove that the kernel is what you expected. (This last step is usually done in two parts, establishing $I = ker(phi)$ via $I subseteq ker(phi)$ and $I supseteqker(phi)$, where $I$ is your desired ideal.)
The other useful hint here is that some of the easiest isomorphisms defined on $R[x]$ are the "point evaluations", i.e. substitute a fixed value for $x$ in each polynomial.
To illustrate this let's work out a very simple example showing that $R[x]/langle x rangle simeq R$ for any ring $R$. Define $phi : R[x] rightarrow R$ by $ phi: p(x) mapsto p(0)$. It's a homomorphism because of the way the ring operations are defined in $R[x]$. It is onto because for any $a in R$ we can define $p(x) = x + a in R[x]$ and $phi(p) = p(0) = a$. If $p(x) = x$ then $phi(p) = 0$ so $langle x rangle subseteq ker(phi)$. To prove the other inclusion, assume $q(x) in ker(phi)$, so $q(0) = 0$, i.e. the constant term of $q$ is zero, so we can factor an $x$ out, i.e. we can write $q$ as $xcdot p(x)$ for some $p$, so $q in langle x rangle$, and this shows $ker(phi) subseteq langle x rangle$. QED.
Notice that when we say we want to take $R[x]$ and mod out by $langle x rangle$ we can think of this as wanting to "set x equal to zero", To test your understanding, if we were given $R[x]/langle x-1 rangle$ we would want to set $x-1$ equal to zero. Which point evaluation would let us do that?
Now in your first problem you want to set $x^2 + x$ equal to zero. It's not quite as simple as the $R[x]/langle xrangle simeq R$ example, but you can think of this as setting $x(x+1)$ equal to zero, so try to combine "setting $x$ equal to zero" and "setting $x+1$ equal to zero" to get a complete answer.
For your second problem you should be able to pick the appropriate point evaluation to handle "setting $x-2$ to zero" as your first step. But then you still have to handle the $x^2 + 1$, or more accurately, whatever $x^2+1$ maps to after you've applied your first homomorphism. Go ahead and figure out what your point evaluation, $phi$, should be, and then tell yourself "Well, I want $phi(x^2+1)$ to be zero", and see if that suggests anything to you.
(I've written this as a bunch of hints to help you get started and guide your thinking, so there are still a bunch of details for you to figure out and nail down. Feel free to follow up with what you figure out and requests if you feel you need further hints.)
answered Jan 2 at 3:36
JonathanZJonathanZ
2,119613
$endgroup$
add a comment |
$begingroup$
I find the easiest setting to think about questions like these are by using the first isomorphism theorem: if $phi$ is a ring homomorphism from $R$ onto $S$, then $S simeq R/ker(phi)$. This means you have to have some suspicion about what the result will look like, define a homomorphism into that ring, show that it is onto, and then prove that the kernel is what you expected. (This last step is usually done in two parts, establishing $I = ker(phi)$ via $I subseteq ker(phi)$ and $I supseteqker(phi)$, where $I$ is your desired ideal.)
The other useful hint here is that some of the easiest isomorphisms defined on $R[x]$ are the "point evaluations", i.e. substitute a fixed value for $x$ in each polynomial.
To illustrate this let's work out a very simple example showing that $R[x]/langle x rangle simeq R$ for any ring $R$. Define $phi : R[x] rightarrow R$ by $ phi: p(x) mapsto p(0)$. It's a homomorphism because of the way the ring operations are defined in $R[x]$. It is onto because for any $a in R$ we can define $p(x) = x + a in R[x]$ and $phi(p) = p(0) = a$. If $p(x) = x$ then $phi(p) = 0$ so $langle x rangle subseteq ker(phi)$. To prove the other inclusion, assume $q(x) in ker(phi)$, so $q(0) = 0$, i.e. the constant term of $q$ is zero, so we can factor an $x$ out, i.e. we can write $q$ as $xcdot p(x)$ for some $p$, so $q in langle x rangle$, and this shows $ker(phi) subseteq langle x rangle$. QED.
Notice that when we say we want to take $R[x]$ and mod out by $langle x rangle$ we can think of this as wanting to "set x equal to zero", To test your understanding, if we were given $R[x]/langle x-1 rangle$ we would want to set $x-1$ equal to zero. Which point evaluation would let us do that?
Now in your first problem you want to set $x^2 + x$ equal to zero. It's not quite as simple as the $R[x]/langle xrangle simeq R$ example, but you can think of this as setting $x(x+1)$ equal to zero, so try to combine "setting $x$ equal to zero" and "setting $x+1$ equal to zero" to get a complete answer.
For your second problem you should be able to pick the appropriate point evaluation to handle "setting $x-2$ to zero" as your first step. But then you still have to handle the $x^2 + 1$, or more accurately, whatever $x^2+1$ maps to after you've applied your first homomorphism. Go ahead and figure out what your point evaluation, $phi$, should be, and then tell yourself "Well, I want $phi(x^2+1)$ to be zero", and see if that suggests anything to you.
(I've written this as a bunch of hints to help you get started and guide your thinking, so there are still a bunch of details for you to figure out and nail down. Feel free to follow up with what you figure out and requests if you feel you need further hints.)
answered Jan 2 at 3:36
JonathanZJonathanZ
2,119613
$endgroup$
I find the easiest setting to think about questions like these are by using the first isomorphism theorem: if $phi$ is a ring homomorphism from $R$ onto $S$, then $S simeq R/ker(phi)$. This means you have to have some suspicion about what the result will look like, define a homomorphism into that ring, show that it is onto, and then prove that the kernel is what you expected. (This last step is usually done in two parts, establishing $I = ker(phi)$ via $I subseteq ker(phi)$ and $I supseteqker(phi)$, where $I$ is your desired ideal.)
The other useful hint here is that some of the easiest isomorphisms defined on $R[x]$ are the "point evaluations", i.e. substitute a fixed value for $x$ in each polynomial.
To illustrate this let's work out a very simple example showing that $R[x]/langle x rangle simeq R$ for any ring $R$. Define $phi : R[x] rightarrow R$ by $ phi: p(x) mapsto p(0)$. It's a homomorphism because of the way the ring operations are defined in $R[x]$. It is onto because for any $a in R$ we can define $p(x) = x + a in R[x]$ and $phi(p) = p(0) = a$. If $p(x) = x$ then $phi(p) = 0$ so $langle x rangle subseteq ker(phi)$. To prove the other inclusion, assume $q(x) in ker(phi)$, so $q(0) = 0$, i.e. the constant term of $q$ is zero, so we can factor an $x$ out, i.e. we can write $q$ as $xcdot p(x)$ for some $p$, so $q in langle x rangle$, and this shows $ker(phi) subseteq langle x rangle$. QED.
Notice that when we say we want to take $R[x]$ and mod out by $langle x rangle$ we can think of this as wanting to "set x equal to zero", To test your understanding, if we were given $R[x]/langle x-1 rangle$ we would want to set $x-1$ equal to zero. Which point evaluation would let us do that?
Now in your first problem you want to set $x^2 + x$ equal to zero. It's not quite as simple as the $R[x]/langle xrangle simeq R$ example, but you can think of this as setting $x(x+1)$ equal to zero, so try to combine "setting $x$ equal to zero" and "setting $x+1$ equal to zero" to get a complete answer.
For your second problem you should be able to pick the appropriate point evaluation to handle "setting $x-2$ to zero" as your first step. But then you still have to handle the $x^2 + 1$, or more accurately, whatever $x^2+1$ maps to after you've applied your first homomorphism. Go ahead and figure out what your point evaluation, $phi$, should be, and then tell yourself "Well, I want $phi(x^2+1)$ to be zero", and see if that suggests anything to you.
(I've written this as a bunch of hints to help you get started and guide your thinking, so there are still a bunch of details for you to figure out and nail down. Feel free to follow up with what you figure out and requests if you feel you need further hints.)
answered Jan 2 at 3:36
JonathanZJonathanZ
2,119613
edited Jan 2 at 4:00
answered Jan 2 at 3:36
JonathanZJonathanZ
2,119613
answered Jan 2 at 3:36
JonathanZJonathanZ
2,119613
answered Jan 2 at 3:36
JonathanZJonathanZ
2,119613
2,119613
add a comment |
add a comment |
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$begingroup$
For example, $(x-2,x^2+1) superseteq (x-2,x^2+1-(x-2)*x)=(x-2,1+4x)$. This is because $(x-2)$ being in the ideal means that $(x-2)*x$ is as well and we can take differences too. If you can use the ideal rules to go backwards as well you can get the other containment. Keep doing this until you get something simpler.
$endgroup$
– AHusain
Jan 2 at 1:53
$begingroup$
For the first one, just think about the definition of quotient ring in general. If $f,ginmathbb{Q}[x]/(x^2+x),$ they are the same (belong to the same equivalence class) iff $x^2+x|f-g.$ In particular, this means that there are only degree $0$ and $1$ polynomials in them. Now, you should also prove that any such polynomial is also in there too
$endgroup$
– dezdichado
Jan 2 at 1:57
$begingroup$
For the first one, factor and use CRT. For the second, note that $x=2$.
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– jgon
Jan 2 at 3:03
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Hey, @AHusain, the LaTeX you want is "supseteq", not "superseteq". (I had to look it up myself.)
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– JonathanZ
Jan 2 at 3:55