Is $S$ is subring of $T$?
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consider $S = C[x^5]$,complex polynomials is $x^5$ , as a subset of $T =C[x]$ the ring of all complex polynomials.
Now My question is that
Is $S$ is subring of $T$ ?
My attempts : i Thinks Yes because $S$ is a subset of $T$
Any hints/solution will be apprecaited
thanks u
abstract-algebra
add a comment |
up vote
0
down vote
favorite
consider $S = C[x^5]$,complex polynomials is $x^5$ , as a subset of $T =C[x]$ the ring of all complex polynomials.
Now My question is that
Is $S$ is subring of $T$ ?
My attempts : i Thinks Yes because $S$ is a subset of $T$
Any hints/solution will be apprecaited
thanks u
abstract-algebra
4
Check also $S$ is closed under subtraction, closed under multiplication, and it contains the multiplicative identity $1$.
– i707107
Nov 19 at 22:27
1
Its a subset but as said, you need to check that it is closed under addition, contains the additive inverse of each element, it is closed under multiplication and contains the unit element 1.
– Wuestenfux
Nov 20 at 12:09
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
consider $S = C[x^5]$,complex polynomials is $x^5$ , as a subset of $T =C[x]$ the ring of all complex polynomials.
Now My question is that
Is $S$ is subring of $T$ ?
My attempts : i Thinks Yes because $S$ is a subset of $T$
Any hints/solution will be apprecaited
thanks u
abstract-algebra
consider $S = C[x^5]$,complex polynomials is $x^5$ , as a subset of $T =C[x]$ the ring of all complex polynomials.
Now My question is that
Is $S$ is subring of $T$ ?
My attempts : i Thinks Yes because $S$ is a subset of $T$
Any hints/solution will be apprecaited
thanks u
abstract-algebra
abstract-algebra
asked Nov 19 at 22:24
jasmine
1,388416
1,388416
4
Check also $S$ is closed under subtraction, closed under multiplication, and it contains the multiplicative identity $1$.
– i707107
Nov 19 at 22:27
1
Its a subset but as said, you need to check that it is closed under addition, contains the additive inverse of each element, it is closed under multiplication and contains the unit element 1.
– Wuestenfux
Nov 20 at 12:09
add a comment |
4
Check also $S$ is closed under subtraction, closed under multiplication, and it contains the multiplicative identity $1$.
– i707107
Nov 19 at 22:27
1
Its a subset but as said, you need to check that it is closed under addition, contains the additive inverse of each element, it is closed under multiplication and contains the unit element 1.
– Wuestenfux
Nov 20 at 12:09
4
4
Check also $S$ is closed under subtraction, closed under multiplication, and it contains the multiplicative identity $1$.
– i707107
Nov 19 at 22:27
Check also $S$ is closed under subtraction, closed under multiplication, and it contains the multiplicative identity $1$.
– i707107
Nov 19 at 22:27
1
1
Its a subset but as said, you need to check that it is closed under addition, contains the additive inverse of each element, it is closed under multiplication and contains the unit element 1.
– Wuestenfux
Nov 20 at 12:09
Its a subset but as said, you need to check that it is closed under addition, contains the additive inverse of each element, it is closed under multiplication and contains the unit element 1.
– Wuestenfux
Nov 20 at 12:09
add a comment |
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4
Check also $S$ is closed under subtraction, closed under multiplication, and it contains the multiplicative identity $1$.
– i707107
Nov 19 at 22:27
1
Its a subset but as said, you need to check that it is closed under addition, contains the additive inverse of each element, it is closed under multiplication and contains the unit element 1.
– Wuestenfux
Nov 20 at 12:09