Reason why $x mapsto x^p$ is a ring morphism in characteristic $p$?
0
Let $R$ be a commutative unital ring of characteristic $p$, where $p$ is prime.
The Frobenius map $x mapsto x^p$ on $R$ is known to be a ring homomorphism (in particular, additive).
The only proof I know of this fact is to use binomial expansion and then $p mid {pchoose k}$ for every $0<k<p$. But what are some other proofs? Are there "conceptual" reason for such a canonical morphism to exist?
commutative-algebra positive-characteristic
asked Nov 29 at 18:33
Alphonse
2,148623
add a comment |
0
Let $R$ be a commutative unital ring of characteristic $p$, where $p$ is prime.
The Frobenius map $x mapsto x^p$ on $R$ is known to be a ring homomorphism (in particular, additive).
The only proof I know of this fact is to use binomial expansion and then $p mid {pchoose k}$ for every $0<k<p$. But what are some other proofs? Are there "conceptual" reason for such a canonical morphism to exist?
commutative-algebra positive-characteristic
asked Nov 29 at 18:33
Alphonse
2,148623
2
Can you clarify why you fin the binomial theorem argument non-conceptual? The only possible barrier to homomorphism-hood is additivity, and the additivity requirement is precisely $(x+y)^p=x^p+y^p$, and it's clear that this would hold if all the nontrivial coefficients in the binomial expansion are multiples of $p$. So if you find this latter fact - "all but the first and last terms in the binomial expansion of $(x+y)^p$ have coefficients which are multiples of $p$" - conceptually clear, you should be happy with the existing proof.
– Noah Schweber
Nov 29 at 19:16
3
Incidentally, here's my favorite proof of that fact. For $Ssubseteq{0,1,..., p-1}$ and $ain {0,1,...,p-1}$, let $S+a={(s+a)mod p: sin S}$. Think of $S+a$ as gotten by "shifting" $S$ over by $a$ (and then modding out appropriately). This defines an equivalence relation on subsets of ${0,1,..., p-1}$: namely, $Ssim T$ iff $S$ is a "shift" of $T$. Now for $0<k<p$, the set of $k$-element subsets of ${0,1,...,p-1}$ is partitioned into $sim$-classes, and each $sim$-class has cardinality $p$ (here's where $0<k<p$ is used), so the number of $k$-element subsets is a multiple of $p$!
– Noah Schweber
Nov 29 at 19:23
add a comment |
0
0
0
Let $R$ be a commutative unital ring of characteristic $p$, where $p$ is prime.
The Frobenius map $x mapsto x^p$ on $R$ is known to be a ring homomorphism (in particular, additive).
The only proof I know of this fact is to use binomial expansion and then $p mid {pchoose k}$ for every $0<k<p$. But what are some other proofs? Are there "conceptual" reason for such a canonical morphism to exist?
commutative-algebra positive-characteristic
asked Nov 29 at 18:33
Alphonse
2,148623
Let $R$ be a commutative unital ring of characteristic $p$, where $p$ is prime.
The Frobenius map $x mapsto x^p$ on $R$ is known to be a ring homomorphism (in particular, additive).
The only proof I know of this fact is to use binomial expansion and then $p mid {pchoose k}$ for every $0<k<p$. But what are some other proofs? Are there "conceptual" reason for such a canonical morphism to exist?
commutative-algebra positive-characteristic
commutative-algebra positive-characteristic
asked Nov 29 at 18:33
Alphonse
2,148623
asked Nov 29 at 18:33
Alphonse
2,148623
asked Nov 29 at 18:33
Alphonse
2,148623
asked Nov 29 at 18:33
Alphonse
2,148623
asked Nov 29 at 18:33
Alphonse
2,148623
2,148623
2
Can you clarify why you fin the binomial theorem argument non-conceptual? The only possible barrier to homomorphism-hood is additivity, and the additivity requirement is precisely $(x+y)^p=x^p+y^p$, and it's clear that this would hold if all the nontrivial coefficients in the binomial expansion are multiples of $p$. So if you find this latter fact - "all but the first and last terms in the binomial expansion of $(x+y)^p$ have coefficients which are multiples of $p$" - conceptually clear, you should be happy with the existing proof.
– Noah Schweber
Nov 29 at 19:16
3
Incidentally, here's my favorite proof of that fact. For $Ssubseteq{0,1,..., p-1}$ and $ain {0,1,...,p-1}$, let $S+a={(s+a)mod p: sin S}$. Think of $S+a$ as gotten by "shifting" $S$ over by $a$ (and then modding out appropriately). This defines an equivalence relation on subsets of ${0,1,..., p-1}$: namely, $Ssim T$ iff $S$ is a "shift" of $T$. Now for $0<k<p$, the set of $k$-element subsets of ${0,1,...,p-1}$ is partitioned into $sim$-classes, and each $sim$-class has cardinality $p$ (here's where $0<k<p$ is used), so the number of $k$-element subsets is a multiple of $p$!
– Noah Schweber
Nov 29 at 19:23
add a comment |
2
Can you clarify why you fin the binomial theorem argument non-conceptual? The only possible barrier to homomorphism-hood is additivity, and the additivity requirement is precisely $(x+y)^p=x^p+y^p$, and it's clear that this would hold if all the nontrivial coefficients in the binomial expansion are multiples of $p$. So if you find this latter fact - "all but the first and last terms in the binomial expansion of $(x+y)^p$ have coefficients which are multiples of $p$" - conceptually clear, you should be happy with the existing proof.
– Noah Schweber
Nov 29 at 19:16
3
Incidentally, here's my favorite proof of that fact. For $Ssubseteq{0,1,..., p-1}$ and $ain {0,1,...,p-1}$, let $S+a={(s+a)mod p: sin S}$. Think of $S+a$ as gotten by "shifting" $S$ over by $a$ (and then modding out appropriately). This defines an equivalence relation on subsets of ${0,1,..., p-1}$: namely, $Ssim T$ iff $S$ is a "shift" of $T$. Now for $0<k<p$, the set of $k$-element subsets of ${0,1,...,p-1}$ is partitioned into $sim$-classes, and each $sim$-class has cardinality $p$ (here's where $0<k<p$ is used), so the number of $k$-element subsets is a multiple of $p$!
– Noah Schweber
Nov 29 at 19:23
2
2
Can you clarify why you fin the binomial theorem argument non-conceptual? The only possible barrier to homomorphism-hood is additivity, and the additivity requirement is precisely $(x+y)^p=x^p+y^p$, and it's clear that this would hold if all the nontrivial coefficients in the binomial expansion are multiples of $p$. So if you find this latter fact - "all but the first and last terms in the binomial expansion of $(x+y)^p$ have coefficients which are multiples of $p$" - conceptually clear, you should be happy with the existing proof.
– Noah Schweber
Nov 29 at 19:16
Can you clarify why you fin the binomial theorem argument non-conceptual? The only possible barrier to homomorphism-hood is additivity, and the additivity requirement is precisely $(x+y)^p=x^p+y^p$, and it's clear that this would hold if all the nontrivial coefficients in the binomial expansion are multiples of $p$. So if you find this latter fact - "all but the first and last terms in the binomial expansion of $(x+y)^p$ have coefficients which are multiples of $p$" - conceptually clear, you should be happy with the existing proof.
– Noah Schweber
Nov 29 at 19:16
3
3
Incidentally, here's my favorite proof of that fact. For $Ssubseteq{0,1,..., p-1}$ and $ain {0,1,...,p-1}$, let $S+a={(s+a)mod p: sin S}$. Think of $S+a$ as gotten by "shifting" $S$ over by $a$ (and then modding out appropriately). This defines an equivalence relation on subsets of ${0,1,..., p-1}$: namely, $Ssim T$ iff $S$ is a "shift" of $T$. Now for $0<k<p$, the set of $k$-element subsets of ${0,1,...,p-1}$ is partitioned into $sim$-classes, and each $sim$-class has cardinality $p$ (here's where $0<k<p$ is used), so the number of $k$-element subsets is a multiple of $p$!
– Noah Schweber
Nov 29 at 19:23
Incidentally, here's my favorite proof of that fact. For $Ssubseteq{0,1,..., p-1}$ and $ain {0,1,...,p-1}$, let $S+a={(s+a)mod p: sin S}$. Think of $S+a$ as gotten by "shifting" $S$ over by $a$ (and then modding out appropriately). This defines an equivalence relation on subsets of ${0,1,..., p-1}$: namely, $Ssim T$ iff $S$ is a "shift" of $T$. Now for $0<k<p$, the set of $k$-element subsets of ${0,1,...,p-1}$ is partitioned into $sim$-classes, and each $sim$-class has cardinality $p$ (here's where $0<k<p$ is used), so the number of $k$-element subsets is a multiple of $p$!
– Noah Schweber
Nov 29 at 19:23
add a comment |
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2
Can you clarify why you fin the binomial theorem argument non-conceptual? The only possible barrier to homomorphism-hood is additivity, and the additivity requirement is precisely $(x+y)^p=x^p+y^p$, and it's clear that this would hold if all the nontrivial coefficients in the binomial expansion are multiples of $p$. So if you find this latter fact - "all but the first and last terms in the binomial expansion of $(x+y)^p$ have coefficients which are multiples of $p$" - conceptually clear, you should be happy with the existing proof.
– Noah Schweber
Nov 29 at 19:16
3
Incidentally, here's my favorite proof of that fact. For $Ssubseteq{0,1,..., p-1}$ and $ain {0,1,...,p-1}$, let $S+a={(s+a)mod p: sin S}$. Think of $S+a$ as gotten by "shifting" $S$ over by $a$ (and then modding out appropriately). This defines an equivalence relation on subsets of ${0,1,..., p-1}$: namely, $Ssim T$ iff $S$ is a "shift" of $T$. Now for $0<k<p$, the set of $k$-element subsets of ${0,1,...,p-1}$ is partitioned into $sim$-classes, and each $sim$-class has cardinality $p$ (here's where $0<k<p$ is used), so the number of $k$-element subsets is a multiple of $p$!
– Noah Schweber
Nov 29 at 19:23