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$I + J = A$ if, and only if, $sqrtI + sqrtJ = A$
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In that notation, $I$ and $J$ are ideals of a commutative ring $A$ and $sqrtI$ denotes the radical of the ideal $I$.
The first implication is obvious. In fact, since $I subset sqrtI$ and $J subset sqrtJ$, we have $A = I + J subset sqrtI + sqrtJ$. This shows that $I + J = A Rightarrow sqrtI + sqrtJ = A $.
For the other side, I need to show that $ sqrtI + sqrtJ = A Rightarrow I + J = A $. I've already demonstrated that $sqrtsqrtI + sqrtJ = sqrtI + J$, then I just have to prove that $sqrtI + J = A Rightarrow I + J = A$, but I don't know how to proceed from here.
abstract-algebra commutative-algebra ideals
asked Jul 28 at 14:21
wesleyla
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In that notation, $I$ and $J$ are ideals of a commutative ring $A$ and $sqrtI$ denotes the radical of the ideal $I$.
The first implication is obvious. In fact, since $I subset sqrtI$ and $J subset sqrtJ$, we have $A = I + J subset sqrtI + sqrtJ$. This shows that $I + J = A Rightarrow sqrtI + sqrtJ = A $.
For the other side, I need to show that $ sqrtI + sqrtJ = A Rightarrow I + J = A $. I've already demonstrated that $sqrtsqrtI + sqrtJ = sqrtI + J$, then I just have to prove that $sqrtI + J = A Rightarrow I + J = A$, but I don't know how to proceed from here.
abstract-algebra commutative-algebra ideals
asked Jul 28 at 14:21
wesleyla
132
1
$1in A=sqrtI+J$ means that there is $n$ such that $1^nin I+J$.
– user578878
Jul 28 at 14:24
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In that notation, $I$ and $J$ are ideals of a commutative ring $A$ and $sqrtI$ denotes the radical of the ideal $I$.
The first implication is obvious. In fact, since $I subset sqrtI$ and $J subset sqrtJ$, we have $A = I + J subset sqrtI + sqrtJ$. This shows that $I + J = A Rightarrow sqrtI + sqrtJ = A $.
For the other side, I need to show that $ sqrtI + sqrtJ = A Rightarrow I + J = A $. I've already demonstrated that $sqrtsqrtI + sqrtJ = sqrtI + J$, then I just have to prove that $sqrtI + J = A Rightarrow I + J = A$, but I don't know how to proceed from here.
abstract-algebra commutative-algebra ideals
asked Jul 28 at 14:21
wesleyla
132
In that notation, $I$ and $J$ are ideals of a commutative ring $A$ and $sqrtI$ denotes the radical of the ideal $I$.
The first implication is obvious. In fact, since $I subset sqrtI$ and $J subset sqrtJ$, we have $A = I + J subset sqrtI + sqrtJ$. This shows that $I + J = A Rightarrow sqrtI + sqrtJ = A $.
For the other side, I need to show that $ sqrtI + sqrtJ = A Rightarrow I + J = A $. I've already demonstrated that $sqrtsqrtI + sqrtJ = sqrtI + J$, then I just have to prove that $sqrtI + J = A Rightarrow I + J = A$, but I don't know how to proceed from here.
abstract-algebra commutative-algebra ideals
asked Jul 28 at 14:21
wesleyla
132
asked Jul 28 at 14:21
wesleyla
132
asked Jul 28 at 14:21
wesleyla
132
asked Jul 28 at 14:21
wesleyla
132
132
1
$1in A=sqrtI+J$ means that there is $n$ such that $1^nin I+J$.
– user578878
Jul 28 at 14:24
add a comment |Â
1
$1in A=sqrtI+J$ means that there is $n$ such that $1^nin I+J$.
– user578878
Jul 28 at 14:24
1
1
$1in A=sqrtI+J$ means that there is $n$ such that $1^nin I+J$.
– user578878
Jul 28 at 14:24
$1in A=sqrtI+J$ means that there is $n$ such that $1^nin I+J$.
– user578878
Jul 28 at 14:24
add a comment |Â
1 Answer
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Assume $sqrt I + sqrt J = A$, i.e. $x + y = 1$ for some $x, y$ where $x^m in I$ and $y^n in J$.
Therefore, $(x+y)^m+n-1 = 1$.
Observe that in the binomial expansion of $(x+y)^m+n-1$, each term is $k x^r y^s$ for some $k in Bbb Z$ and $r+s = m+n-1$. If $r < m$ and $s < n$, then $r le m - 1$ and $s le n - 1$, so $r + s le m + n - 2$, contradiction; therefore either $r ge m$ or $s ge n$. Therefore, the term either belongs to $I$ or belongs to $J$. Therefore, $1 = (x+y)^m+n-1 in I + J$.
This is basically the proof that the radical of an ideal is an ideal.
answered Jul 28 at 14:29
Kenny Lau
18.1k2156
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Assume $sqrt I + sqrt J = A$, i.e. $x + y = 1$ for some $x, y$ where $x^m in I$ and $y^n in J$.
Therefore, $(x+y)^m+n-1 = 1$.
Observe that in the binomial expansion of $(x+y)^m+n-1$, each term is $k x^r y^s$ for some $k in Bbb Z$ and $r+s = m+n-1$. If $r < m$ and $s < n$, then $r le m - 1$ and $s le n - 1$, so $r + s le m + n - 2$, contradiction; therefore either $r ge m$ or $s ge n$. Therefore, the term either belongs to $I$ or belongs to $J$. Therefore, $1 = (x+y)^m+n-1 in I + J$.
This is basically the proof that the radical of an ideal is an ideal.
answered Jul 28 at 14:29
Kenny Lau
18.1k2156
add a comment |Â
up vote
2
down vote
accepted
Assume $sqrt I + sqrt J = A$, i.e. $x + y = 1$ for some $x, y$ where $x^m in I$ and $y^n in J$.
Therefore, $(x+y)^m+n-1 = 1$.
Observe that in the binomial expansion of $(x+y)^m+n-1$, each term is $k x^r y^s$ for some $k in Bbb Z$ and $r+s = m+n-1$. If $r < m$ and $s < n$, then $r le m - 1$ and $s le n - 1$, so $r + s le m + n - 2$, contradiction; therefore either $r ge m$ or $s ge n$. Therefore, the term either belongs to $I$ or belongs to $J$. Therefore, $1 = (x+y)^m+n-1 in I + J$.
This is basically the proof that the radical of an ideal is an ideal.
answered Jul 28 at 14:29
Kenny Lau
18.1k2156
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Assume $sqrt I + sqrt J = A$, i.e. $x + y = 1$ for some $x, y$ where $x^m in I$ and $y^n in J$.
Therefore, $(x+y)^m+n-1 = 1$.
Observe that in the binomial expansion of $(x+y)^m+n-1$, each term is $k x^r y^s$ for some $k in Bbb Z$ and $r+s = m+n-1$. If $r < m$ and $s < n$, then $r le m - 1$ and $s le n - 1$, so $r + s le m + n - 2$, contradiction; therefore either $r ge m$ or $s ge n$. Therefore, the term either belongs to $I$ or belongs to $J$. Therefore, $1 = (x+y)^m+n-1 in I + J$.
This is basically the proof that the radical of an ideal is an ideal.
answered Jul 28 at 14:29
Kenny Lau
18.1k2156
Assume $sqrt I + sqrt J = A$, i.e. $x + y = 1$ for some $x, y$ where $x^m in I$ and $y^n in J$.
Therefore, $(x+y)^m+n-1 = 1$.
Observe that in the binomial expansion of $(x+y)^m+n-1$, each term is $k x^r y^s$ for some $k in Bbb Z$ and $r+s = m+n-1$. If $r < m$ and $s < n$, then $r le m - 1$ and $s le n - 1$, so $r + s le m + n - 2$, contradiction; therefore either $r ge m$ or $s ge n$. Therefore, the term either belongs to $I$ or belongs to $J$. Therefore, $1 = (x+y)^m+n-1 in I + J$.
This is basically the proof that the radical of an ideal is an ideal.
answered Jul 28 at 14:29
Kenny Lau
18.1k2156
answered Jul 28 at 14:29
Kenny Lau
18.1k2156
answered Jul 28 at 14:29
Kenny Lau
18.1k2156
answered Jul 28 at 14:29
Kenny Lau
18.1k2156
18.1k2156
add a comment |Â
add a comment |Â
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1
$1in A=sqrtI+J$ means that there is $n$ such that $1^nin I+J$.
– user578878
Jul 28 at 14:24