Checking surjectivity of a map.
Let $R = mathbb{R}[x,y]$ be bivariate polynomial ring and $I_1 = langle f_1 rangle$, $I_2 = langle f_2 rangle$ and $I_3 = langle f_3 rangle$ be three principal ideals of $R$. Consider the following map.
$alpha : R^3 to dfrac{R}{I_1} oplus dfrac{R}{I_2} oplus dfrac{R}{I_3}$, with the following rule
$alpha (a,b,c) = (a - b + I_1 , b - c + I_2 , a- c + I_3)$.
My question is that, how can I check whether $alpha$ is surjective or not? What I've done is that, let $(bar{p} , bar{q} , bar{r}) in dfrac{R}{I_1} oplus dfrac{R}{I_2} oplus dfrac{R}{I_3}$ where $bar{p} = p + I_1$, $bar{q} = q + I_2$ , $bar{r} = r + I_3$. So can I find such $a,b,c in R$ that $a - b - p in I_1$ , $b - c - q in I_2$ and $a - c - r in I_3$? Here is where I stuck.
Also how can $f_1 , f_2$ and $f_3$ effect the surjectivity of $alpha$ ?
abstract-algebra polynomials
asked Jan 4 at 8:25
Philip JohnsonPhilip Johnson
91
add a comment |
Let $R = mathbb{R}[x,y]$ be bivariate polynomial ring and $I_1 = langle f_1 rangle$, $I_2 = langle f_2 rangle$ and $I_3 = langle f_3 rangle$ be three principal ideals of $R$. Consider the following map.
$alpha : R^3 to dfrac{R}{I_1} oplus dfrac{R}{I_2} oplus dfrac{R}{I_3}$, with the following rule
$alpha (a,b,c) = (a - b + I_1 , b - c + I_2 , a- c + I_3)$.
My question is that, how can I check whether $alpha$ is surjective or not? What I've done is that, let $(bar{p} , bar{q} , bar{r}) in dfrac{R}{I_1} oplus dfrac{R}{I_2} oplus dfrac{R}{I_3}$ where $bar{p} = p + I_1$, $bar{q} = q + I_2$ , $bar{r} = r + I_3$. So can I find such $a,b,c in R$ that $a - b - p in I_1$ , $b - c - q in I_2$ and $a - c - r in I_3$? Here is where I stuck.
Also how can $f_1 , f_2$ and $f_3$ effect the surjectivity of $alpha$ ?
abstract-algebra polynomials
asked Jan 4 at 8:25
Philip JohnsonPhilip Johnson
91
$f_1, f_2, f_3$ affect the surjectivity of $alpha$. Take for example the trivial case $f_1=f_2=f_3=1$.
– mathcounterexamples.net
Jan 4 at 8:35
@mathcounterexamples.net actually $1$ generates whole ring and the statement becomes trivially true. That's an interesting question.
– freakish
Jan 4 at 9:17
add a comment |
Let $R = mathbb{R}[x,y]$ be bivariate polynomial ring and $I_1 = langle f_1 rangle$, $I_2 = langle f_2 rangle$ and $I_3 = langle f_3 rangle$ be three principal ideals of $R$. Consider the following map.
$alpha : R^3 to dfrac{R}{I_1} oplus dfrac{R}{I_2} oplus dfrac{R}{I_3}$, with the following rule
$alpha (a,b,c) = (a - b + I_1 , b - c + I_2 , a- c + I_3)$.
My question is that, how can I check whether $alpha$ is surjective or not? What I've done is that, let $(bar{p} , bar{q} , bar{r}) in dfrac{R}{I_1} oplus dfrac{R}{I_2} oplus dfrac{R}{I_3}$ where $bar{p} = p + I_1$, $bar{q} = q + I_2$ , $bar{r} = r + I_3$. So can I find such $a,b,c in R$ that $a - b - p in I_1$ , $b - c - q in I_2$ and $a - c - r in I_3$? Here is where I stuck.
Also how can $f_1 , f_2$ and $f_3$ effect the surjectivity of $alpha$ ?
abstract-algebra polynomials
asked Jan 4 at 8:25
Philip JohnsonPhilip Johnson
91
Let $R = mathbb{R}[x,y]$ be bivariate polynomial ring and $I_1 = langle f_1 rangle$, $I_2 = langle f_2 rangle$ and $I_3 = langle f_3 rangle$ be three principal ideals of $R$. Consider the following map.
$alpha : R^3 to dfrac{R}{I_1} oplus dfrac{R}{I_2} oplus dfrac{R}{I_3}$, with the following rule
$alpha (a,b,c) = (a - b + I_1 , b - c + I_2 , a- c + I_3)$.
My question is that, how can I check whether $alpha$ is surjective or not? What I've done is that, let $(bar{p} , bar{q} , bar{r}) in dfrac{R}{I_1} oplus dfrac{R}{I_2} oplus dfrac{R}{I_3}$ where $bar{p} = p + I_1$, $bar{q} = q + I_2$ , $bar{r} = r + I_3$. So can I find such $a,b,c in R$ that $a - b - p in I_1$ , $b - c - q in I_2$ and $a - c - r in I_3$? Here is where I stuck.
Also how can $f_1 , f_2$ and $f_3$ effect the surjectivity of $alpha$ ?
abstract-algebra polynomials
abstract-algebra polynomials
asked Jan 4 at 8:25
Philip JohnsonPhilip Johnson
91
asked Jan 4 at 8:25
Philip JohnsonPhilip Johnson
91
asked Jan 4 at 8:25
Philip JohnsonPhilip Johnson
91
asked Jan 4 at 8:25
Philip JohnsonPhilip Johnson
91
asked Jan 4 at 8:25
Philip JohnsonPhilip Johnson
91
91
$f_1, f_2, f_3$ affect the surjectivity of $alpha$. Take for example the trivial case $f_1=f_2=f_3=1$.
– mathcounterexamples.net
Jan 4 at 8:35
@mathcounterexamples.net actually $1$ generates whole ring and the statement becomes trivially true. That's an interesting question.
– freakish
Jan 4 at 9:17
add a comment |
$f_1, f_2, f_3$ affect the surjectivity of $alpha$. Take for example the trivial case $f_1=f_2=f_3=1$.
– mathcounterexamples.net
Jan 4 at 8:35
@mathcounterexamples.net actually $1$ generates whole ring and the statement becomes trivially true. That's an interesting question.
– freakish
Jan 4 at 9:17
$f_1, f_2, f_3$ affect the surjectivity of $alpha$. Take for example the trivial case $f_1=f_2=f_3=1$.
– mathcounterexamples.net
Jan 4 at 8:35
$f_1, f_2, f_3$ affect the surjectivity of $alpha$. Take for example the trivial case $f_1=f_2=f_3=1$.
– mathcounterexamples.net
Jan 4 at 8:35
@mathcounterexamples.net actually $1$ generates whole ring and the statement becomes trivially true. That's an interesting question.
– freakish
Jan 4 at 9:17
@mathcounterexamples.net actually $1$ generates whole ring and the statement becomes trivially true. That's an interesting question.
– freakish
Jan 4 at 9:17
add a comment |
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$f_1, f_2, f_3$ affect the surjectivity of $alpha$. Take for example the trivial case $f_1=f_2=f_3=1$.
– mathcounterexamples.net
Jan 4 at 8:35
@mathcounterexamples.net actually $1$ generates whole ring and the statement becomes trivially true. That's an interesting question.
– freakish
Jan 4 at 9:17