Checking surjectivity of a map.












1














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 $alpha$ ?










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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


















1














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 $alpha$ ?










share|cite|improve this question






















  • $f_1, f_2, f_3$ 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
















1












1








1







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 $alpha$ ?










share|cite|improve this question













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






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asked Jan 4 at 8:25









Philip JohnsonPhilip Johnson

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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




















  • $f_1, f_2, f_3$ 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












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