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7

For a polynomial of degree $3$, if it's not irreducible then it must split in factors that have degree $2$ or $1$, and there must be a factor of degree $1$ (so a root). Hence it suffices to check that there are roots. In degree $5$, your polynomial can split in two factors of respective degree $2$ and $3$ : so you can be reducible without having roots.


6

They are not isomorphic as fields: While $\mathbb{Q}(i \sqrt{5})$ contains a square root of $-5$ (namely $i \sqrt{5}$), the field $\mathbb{Q}(\sqrt{5})$ does not (as it is a subfield of $\mathbb{R}$). They are however isomorphic as $\mathbb{Q}$-vector spaces, as they are both two-dimensional over $\mathbb{Q}$ (because $\mathbb{Q}(\sqrt{5})$ has $\{1, ...


4

It's impossible: let $\mathfrak m$ be a maximal ideal of $A$ which doesn't contain $\ker f$. As $B\simeq A/\ker f$, we have $$f(\mathfrak m)=\mathfrak m \cdot A/\ker f=(\mathfrak m+\ker f)/\ker f=A/\ker f,$$ which is not a prime ideal in the quotient.


4

Start with the map $$\begin{align}\Bbb{Z}_3\left[x\right]&\to \Bbb{Z}_3\left[i\right]\\P&\to P(i)\end{align}$$ It is a ring morphism by definition of the substitution that is surjective. Indeed take any $a+bi\in \Bbb{Z}_3\left[i\right]$ it is the image of the polynomial $bx+a$. The kernel of this morphism is the ideal generated by $x^2+1$ ...


4

There are plenty of cases where this happens. It's really only interesting when the module action interacts in an interesting way with the multiplication The most straightforward example, which works when $R$ is commutative, is the concept of an $R$-algebra, which I'll let you look up yourself. In this case it's easy to see what the action does. If you ...


4

To find $h(X)$: $-3$ is a root of both $f(X)$ and $g(X)$ so $X+3$ divides both $f(X)$ and $g(X)$. Also you can write $x+3$ as a linear combination of $f(X)$ and $g(X)$. So let $h(X)=X+3$ then $(f)+(g)=(h)$. Note that $h$ is the greatest common divisor of $f$ and $g$. To find $k$, factor $f$ and $g$, then you get $f(X)=h(X)f_1(X)$ and $g(X)=h(X)g_1(X)$ (with ...


3

Suppose we have a ring homomorphism $f\colon R\to\mathbb{Z}$. The composite of the homomorphism $f$ following the homomorphism $j\colon \mathbb{Z}\to R : n\mapsto n\cdot 1_R$ is the identity homomorphism $f\circ j=\mathrm{id}_{\mathbb{Z}}$ (this because $\mathrm{id}_{\mathbb{Z}}$ is the only endomorphism of the ring $\mathbb{Z}$), which implies that $j$ is ...


3

If $p = 2$ then $-1$ is a square mod $p$, so we may now assume that $p$ is odd. Suppose that $n^2 \equiv -1 \bmod p$ for some $n \in \mathbb{N}$. Then, using that $\mathbb{Z}/p\mathbb{Z}$ is a field, $$ 1 \equiv n^{p-1} \equiv (n^2)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p - 1}{2}} \bmod p $$ which implies that $\frac{p-1}{2}$ is even, thus, that $p \equiv 1 ...


3

For $p>2$ prime, $-1$ is a quadratic residue modulo $p$ if and only if $p \equiv 1 \bmod 4$. So you can determine whether $-1$ is a quadratic residue modulo $p$ as quickly as you can find the remainder of $p$ upon division by $4$.


3

You are right about $$R=\left\{\begin{bmatrix}\alpha & \beta \\ \bar\beta & \bar\alpha\end{bmatrix}\mid \alpha,\beta\in\mathbb{C}\right\} $$, especially since $e=\frac12\begin{bmatrix}1&1\\ 1&1\end{bmatrix}$ would create zero divisors : $e(1-e)=0$. But presumably what you're reading is about $$R=\left\{\begin{bmatrix}\alpha & \beta ...


3

A finite dimensional algebra will be Artinian, and the prime ideals must be maximal. Therefore the Jacobson and nilradicals coincide.


3

No. If $k\ge 2$, then the quotient will have zero divisors and not even be an integral domain. (A PID needs to be an integral domain by definition).


2

The answer is no. For example, let $y$ be an element of $F$ and let $g(x) = (x-y)^2f(x)$, where $f(x)$ is in $F[x]$ and has all simple roots. Then the splitting field of $g$ over $F$ is the same as that of $f$.


2

For any $r\in R,$ and any $a\in I,$ we have $r.a\in I.$ Take $a=1.$ Then $r.1=r \in I;$ For any $r\in R.$


2

Every subring of a product of fields is also von Neumann regular, so you can pick a subring that isn't a product of fields. For example, choose $F$ to be the field of two elements, and let $R$ be the subring of $\prod_{i=1}^\infty F$ generated by the identity and $\oplus_{i=1}^\infty F$. So $R$ is a von Neumann regular ring, but it is not a product of ...


2

The morphism $x\rightarrow t, y\rightarrow t^2$, $z\rightarrow 1$ induces a morphism You have $f:k[x,y,z)/(x^2-yz,z-1)\rightarrow k[t]$ defined by $f([x])=t, f([y])=t^2, f([z])=1$ where $[x]$ is the class of $x$ Consider $g:k[t]\rightarrow k[x,y,z]/(x^2-yz,z-1)$ defined by $g(t)=[x]$, you have $f(g(t)=f([x])=t$. $g(f([x]))=g(t)=[x]$, ...


2

Every finite dimensional $k$-algebra is isomorphic to a subalgebra of $M_n(k)$ for some $n$ (you can take $n=\dim A$, in fact) So no, for then all algebras would be semisimple!


2

Let $I=\cup_{n\geq 1}I_n$, and observe that $I$ is an ideal. (While it is not true in general that the union of a collection of ideals is an ideal, it is true for an ascending union). Since $A$ is a PID, $I$ can be generated by some element $x$. Let $n$ be the minimal index for which $x\in I_n$, then we have $I_n=I$, hence $I_n=I_{n+k}$ for all $k\geq 0$.


2

It's actually really easy. I've thought a solution through: Let $R$ be a Noetherian ring. If it is uniform, there's nothing to prove. If it is not uniform, then there is a (non-trivial) direct sum of ideals contained in $R$, say $R_{1} \oplus R_{1}'$. So, now, either $R_{1}$ is uniform or there is a (nontrivial) direct sum of ideals in $R_{1}$, say $R_{2} ...


1

Suppose $r,s\in R$ and $rs\in f(P)$. Since $f$ is surjective, you have $r=f(a)$ and $s=f(b)$ for some $a,b\in A$, so the condition is $$ f(a)f(b)=f(c) $$ for some $c\in P$. In turn this becomes $ab-c\in\ker f$. If $P\supseteq\ker f$, you can conclude $ab-c\in P$, so also $ab\in P$ and it's easy to finish. Without the assumption $\ker f\subseteq P$ there's ...


1

Hint: This is primarily a counterexample that shows the kernel to be non-trivial, but it also serves as a strong hint. Let $f(x) = x^2 - 3x + 2$. Then $\varphi(f(x)) = f(1) = 1 - 3 + 2 = 0 \implies f(x) \in \ker \varphi$. Solution: \begin{align*} \ker \varphi & = \{\, f(x) \in \mathbb R[x] \mid \varphi(f(x)) = 0 \in \mathbb R \,\}\\ & = \{\, f(x) ...


1

You are basically correct. If you find a homomorphism between the rings, you only need to show if that particular homomorphism is bijective. (This approach is quite standard and in no way restricted to your particular case.) The first isomorphism theorem is now trivial, so it can be left out.


1

Suppose that $p$ is in the kernel. Then by polynomial division you can write $p=q(X^2-Y^3)+r$ where no term of $r$ has degree $\ge 2$ in $X$, and clearly $r$ is then also in the kernel. Now $r$ consists of terms of the form $aY^n$ and terms of the form $aXY^n$. Under $f$ the former terms become $at^{2n}$ and the latter terms become $at^{2n+3}$. But $t^{2n}$ ...


1

Hint: Ideals are closed under external multiplication. For all $r \in F$, $x \in I$, $rx \in I$. Just apply this to the elements you obtained. Complete Solution: Let $I$ be any non-zero ideal of a field $F$. Then, $I$ contains at least one non-zero element $x$. Since $F$ is a field, $x^{-1} \in F$, which implies that $x^{-1} x = 1 \in I$ (since $I$ is an ...


1

Let $p \in \mathbb{Z}$ be a prime integer and consider $\mathbb{Z}_p$, the ring of $p$-adic integers. Then $R$ has prime ideals $(0)$ and $(p)$, and hence is a discrete valuation ring. It is easy to check that the localization of $\mathbb{Z}_p$ at each prime ideal is a regular local ring, verifying that $\mathbb{Z}_p$ is a regular commutative ring. Since ...


1

Suppose $I$ is an ideal of $\mathbb{R}[x]$ properly containing $\langle x^2+1\rangle$ and let $f(x)\in I\setminus\langle x^2+1\rangle$. Then $$ f(x)=q(x)(x^2+1)+ax+b $$ for some $q(x)\in\mathbb{R}[x]$, $a,b\in\mathbb{R}$. Since $f(x)\notin\langle x^2+1\rangle$, you have $ax+b\ne0$, that is, not both $a$ and $b$ are zero. If $a=0$, then $b\ne0$ and so ...


1

Hint: $\mathbf R[x]$ is a P.I.D.. In a P.I.D., the ideal generated by an irreducible element is maximal.


1

Hint: $R/I$ is isomorphic to the ring $S$ of $2\times2$ matrices with coefficients in $\mathbb{Z}/2\mathbb{Z}$. Can you find a surjective ring homomorphism $R\to S$? Can you tell what its kernel is?


1

The upper triangular matrices form a subalgebra of $M_2(F)$, and they are an obvious counterexample to the proposition.


1

The absorbing (or zero) element of $A/aA$ is $aA$. Therefore if $b+aA=aA$ you just get the trivial equality $0.(c+aA)=0$ in $A/aA$.



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