6

This is a bad axiom and you should not have it. The zero ring is a perfectly fine ring, and without it the category of rings fails to have some very natural constructions, such as tensor products. In algebraic geometry this would mean that affine schemes would fail to be closed under fiber products which would be very silly. You want to be able to talk about ...


6

If the ring $R$ has characteristic $p^2$, then it must be $\mathbb Z/(p^2)$. Otherwise $R$ has characteristic $p$, so there’s a ring homomorphism $\mathbb F_p\to R$ (and so $R$ is an $\mathbb F_p$-algebra). Take any element $x$ not in the image of $\mathbb F_p$. Then there is an algebra homomorphism $\mathbb F_p[t]\to R$, $t\mapsto x$, necessarily surjective ...


5

$$\begin{split} B(A+B)^{-1}A&=B(A+B)^{-1}(A+B-B)\\&= B-B(A+B)^{-1}B\\&=B-(A+B-A)(A+B)^{-1}B \\&= B - B + A(A+B)^{-1}B\\& = A(A+B)^{-1}B \end{split}$$


3

In a ring with $1$ the assumption $1\ne0$ is not necessary and, in fact, the standard modern definition allows the zero ring to exist. The way I personally make sense of this is that the zero ring is the terminal object of the category of rings with $1$, and the reason rings with $1$ should have a terminal object is that they naturally have products of every ...


3

Yes, there is a general way of checking irreducibility of polynomials over finite fields. It is called the Berlekamp algorithm and is based on linear algebra (see your link, which it the Berlekamp–Welch algorithm). For the above polynomial is gives $$x^5 + 2x^4 + x^2 + 2x + 1=(x^3 + x^2 + 2)(x^2 + x + 2).$$ Your computation is the right way to solve the ...


3

More fun if $R$ is not assumed unital. $p$ is prime. I think we get the following cases: If $char(R) = p^2$ (the least integer such that $\forall r\in R,nr=0$) we get one of $\Bbb{Z}/p^2\Bbb{Z},p\Bbb{Z}/p^3\Bbb{Z},p^2\Bbb{Z}/p^4\Bbb{Z}$. Otherwise $char(R)=p$ so $R$ is a two dimensional vector space $R=a\Bbb{F}_p+b\Bbb{F}_p$. Let $Ann(R) = \{ r\in R, \...


3

Hint: $S^{-1}A = \{\frac{a}{b^n}\mid a\in A, n\geq 0\}$. The quotient ring $Q = A[x]/(bX-1)$ contains a zero of $bX-1$, namely the residue class of $x$, which you can denote by $b^{-1}$. This gives the ring extension $A[b^{-1}]$.


3

The entire argument is hidden in the step $(BA)^T = A^T B^T$ and the notation obscures what's going on slightly; on the LHS the matrices are being multiplied in $M_n(R)$ and on the RHS the matrices are being multiplied in $M_n(R^{op})$ (and the statement is false if $R$ is noncommutative and both matrix multiplications are interpreted in the same matrix ring)...


3

$I=(x)$ and $A=(x,y)$ work. In fact, let $J$ be such that $JA=I$. We have $$JA=xJ+yJ=(x)$$ and therefore $x\mid yf$ for all $f\in J$. This means that $x\mid f$ for all $f\in J$ and, therefore, that $J\subseteq (x)$. If $J\subsetneq (x)$, then $JA\subseteq J\subsetneq (x)$: hence, $J=(x)$ is necessary. But $IA=(x^2,xy)\ne (x)$, because, for instance, if $f\in ...


3

The number field $K=\Bbb Q(\sqrt{5})$ has class number one because its Minkowski bound satisfies $B_K<2$ . Hence its ring of integers $\mathcal{O}_K=\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is even a PID and hence a UFD. On the other hand, it is enough to see that $\mathcal{O}_K/(4+\sqrt{5})$ is a field, so that the ideal $(4+\sqrt{5})$ is prime.


3

Call $A = \mathbb Z \left[ \frac{1 + \sqrt 5}2\right]$. We can show that $A / (4+\sqrt 5) \cong \mathbb Z/11 \mathbb Z$, so that the ideal $(4 + \sqrt 5)$ is maximal. As $N(4 + \sqrt 5) = 11$, it is clear that the elements $0, 1, \ldots, 10$ are pairwise incongruent modulo $4 + \sqrt 5$. Every element of $A$ is congruent to an integer modulo $4 + \sqrt 5$: ...


3

Note that $11=(4+\sqrt{5})(4-\sqrt{5})\in\langle 4+\sqrt{5}\rangle$, and so we have the chain of isomorphisms $$\frac{\mathbb{Z}[\sqrt{5}]}{\langle4+\sqrt{5}\rangle}=\frac{\mathbb{Z}[\sqrt{5}]}{\langle4+\sqrt{5},11\rangle}\cong\frac{\mathbb{Z}[t]}{\langle t^2-5,4+t,11\rangle}.$$ Also, $t^2-5=11-(4-t)(4+t)$, whence $\langle t^2-5,4+t,11\rangle=\langle 4+t,11\...


2

If you reflect on this, you will start to see that this is true by definition of prime ideals: They are exactly those ideals into which you cannot multiply from outside. Whether you state it as \begin{align*} xy ∈ \mathfrak p &\implies& x ∈ \mathfrak p &~~~\text{or}~~~ y ∈ \mathfrak p, &\quad \text{or as} \\ xy ∈ R \setminus \mathfrak p &...


2

It is not a necessary axiom, but, as far as I know, there is no benefit in allowing such a thing. The additional structure that you would get is quite boring. Assuming that $0=1$ in $R$, we get that, for every element $a \in R$, $$a=1 \cdot a = 0 \cdot a=0,$$ so $R$ is a ring of only one element.


2

There are only three irreducible polynomials of degree $2$ in $\mathbb{Z}_3[x]$: $$ x^2 + 1, \quad x^2 + x + 2, \quad x^2 + 2 x + 2 $$ Try to divide your polynomial by each of these.


2

Induction helps because if $(x+y)^{p^k}=x^{p^k}+y^{p^k}$ then$$(x+y)^{p^{k+1}}=(x^{p^k}+y^{p^k})^p=(x^{p^k})^p+(y^{p^k})^p,$$provided we've proven the base step $m=1$. @J.W.Tanner hinted checking the base step with$$(x+y)^p-x^p-y^p=\sum_{j=1}^{p-1}\binom{p}{j}x^jy^{p-j}.$$Each binomial coefficient has a factor of $p$ in its numerator $p!$, but not its ...


2

Your proof of the fact that it is a subring is fine. And it is not contained in the center of $\Bbb H$ because $i\in\Bbb R[i]$ and $ij=-ji\ne ji$.


2

So first of all, I want to clear up the definition a bit. You said: Since the set function $i:X \longrightarrow F[X]$ is unknown, it is hard for me to construct the set function $g:X \longrightarrow A$ in order to get a contradiction. The function $i$ is actually not unknown. The issue here is a slight abuse of notation in the problem statement. You are ...


2

You have an injective homomorphism from $A'=A/\mathfrak m\cap A\hookrightarrow B'=B/\mathfrak m$. You can use the following lemma: Let $B$ an integral domain, $A$ a subring of $B$ such that $B$ is a finite $A$-module. Then $B$ is a field if & only if $A$ is a field.


2

You might consider that $M_n(R)$ is (isomorphic to) the ring of endomorphisms of $R_R^n$ (right $R$-module). Thus $M_n(R^{\mathrm{op}})$ is the ring of endomorphisms of $(R^{\mathrm{op}})^n_{R^{\mathrm{op}}}$, which is the same as the ring of endomorphisms of $_RR^{n}$ (left $R$-module), which is $M_n(R)^{\mathrm{op}}$.


2

Yes, $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$ is a UFD because it is norm-Euclidean.


1

$A/((X)/(X^3))=(\mathbb{Q}[X]/(X^3))/((X^2)/(X^3))\cong \mathbb{Q}[X]/(X^2)$. Now $X\neq 0\in \mathbb{Q}[X]/(X^2)$ and $X\cdot X=0\in \mathbb{Q}[X]/(X^2)$, so $\mathbb{Q}[X]/(X^2)$ is not integral domain.


1

Hint: As with many statements like this, it's useful to proceed by way of contradiction. Assume that $x,y\in S$, but $xy\not\in S$. What does this mean about inclusions in $\mathfrak p$? Can you use that to show that this contradicts the definition of prime ideal?


1

Any function $x\mapsto x^k$ with arbitrary $k$ preserves multiplication in a commutative ring. And, use the binomial theorem to prove $x\mapsto x^p$ preserves addition in a ring with characteristic $p$, where $p$ is a prime (specifically $3$ in your example).


1

A straightforward way to prove it using induction is to assume that it holds for all $m$ such that $1\le m\le n$ and for some given $n$. To show that it holds for $n+1$, we should prove $$ (x+y)^{p^{n+1}}=x^{^{p^{n+1}}}+y^{^{p^{n+1}}} $$ proof Note that $$ (x+y)^{p^{n+1}}{= \left[(x+y)^{p^{n}}\right]^p\\ =\left[x^{p^n}+y^{p^n}\right]^p\\ =(x^{p^n})^p+(y^{p^n}...


1

Hint: Use the first isomorphism theorem for the natural map $R[x]\to R/I$ sending $x\mapsto 0$.


1

It's critical here that the ring is finite. Indeed, let $a \in R$ be nonzero. If $a$ is not a zero divisor then the map $R \longrightarrow R$ via $x \mapsto ax$ is injective. As $R$ is finite, it is therefore a bijection by the Pigeonhole Principle. Hence, there is some $x \in R$ such that $ax = 1$, i.e. $a$ is a unit. Note that this is false in general for ...


1

Yes. For instance, let $A=k[x,y]/(x,y)^5$. For any $c\in k\setminus\{0,1\}$, consider the algebra $B_c=A/(xy(x+y)(cx+y))$. Write $V_c=m/m^2$ where $m=(x,y)\subset B_c$ is the unique maximal ideal. We can identify $V_c$ with the set of homogeneous linear polynomials in $x$ and $y$, and thus identify its projectivization with $\mathbb{P}^1_k$. Now note ...


1

We have found $\nu\colon N\to R$ with image $(a_1)$, and taken $y\in N$ with $\nu(y)=a_1$. Given $n\in N$ we have $\nu(n)\in(a_1)$, say $\nu(n)=ta_1$, so that $$ \nu(n)y_1 = ta_1y_1 = ty \in N. $$ Thus both $n$ and $\nu(n)y_1$ lie in $N$, so their difference $n-\nu(n)y_1$ does too. In general a decomposition $M=Rm\oplus K$ will not pass to the submodule $N$, ...


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